KINETIC    THEORY 


OF 


ENGINEERING  STRUCTURES 


DEALING    WITH 


STRESSES,  DEFORMATIONS  AND  WORK 


FOR    THE   USE    OF 


STUDENTS  AND  PRACTITIONERS  IN    CIVIL   ENGINEERING 


BY 

DAVID   A.    MOLITOR,   B.C.E.,  C.E. 

» < 

AUTHOR  "HYDRAULICS  OF  RIVERS,  WEIRS  AND  SLUICES,"  ETC.;  MEM.  AM.  soc.  c.  E.;  MEM.  DETROIT 

ENGINEERING    SOC.;    MEM.    SOC.    FOR   THE    PROM.    ENG.    EDU.J     MEM.    A.  A.  A.  S.,   ETC.; 

FORMERLY    DESIGNING     ENGINEER    ISTHMIAN     CANAL    COMMISSION; 

FROFESSOr,  IN  CIVIL  ENGINEERING,  CORNELL  UNIVERSITY. 


McGRAW-HILL    BOOK    COMPANY 

239   WEST   39TH   STREET,   NEW   YORK 

6  BOUVERIE  STREET,  LONDON,  E.G. 

1911 


Copyright,  1910, 

BY 
DAVID  A.  MOLITOR 


THE   SCIENTIFIC    PHES9 
ROBERT  ORUMMONO   AND  COMP 
BROOKLYN.   N.  V. 


PREFACE 


"  ALL  that  mankind  has  done,  thought,  gained,  or  been;  it  is  lying  in  magic  preserva- 
tion in  the  pages  of  books";  and  so  in  presenting  this  work  for  publication,  the  author 
hopes  to  preserve  the  results  of  many  years  of  painstaking  labors,  study  and  exped- 
ience, to  civil  engineering,  his  chosen  profession. 

The  field  of  usefulness  which  it  is  proposed  to  supply  is  that  of  an  advanced  treatise 
on  stresses  and  deformations  in  engineering  structures,  for  the  use  of  students  and  prac- 
titioners in  civil  engineering  and  particularly  for  specialists  engaged  in  the  design  of 
so-called  "  higher  structures "  requiring  more  than  the  ordinary  methods  of  statics  for 
their  analysis. 

The  reader  is  supposed  to  possess  a  thorough  knowledge  of  higher  mathematics, 
including  the  calculus;  also,  of  the  elements  of  statics,  embracing  the  composition 
and  resolution  of  forces,  force  and  equilibrium  polygons,  etc.  All  these  are  usually 
given  in  the  two  first  years  of  a  Civil  Engineering  course  in  connection  with  mathema- 
tics and  mechanics. 

Details  pertaining  to  the  design  of  bridge  members  and  connections  are  not  dealt 
with. 

The  present  volume  might  advantageously  be  employed  as  a  text-book  in  bridge 
stresses  with  the  aim  of  giving  a  more  thorough  training  in  the  analysis  of  stresses  and 
deformations  by  economizing  somewhat  on  the  time  at  present  devoted  to  detailing 
and  shop  practice,  which  is  not  really  justified  in  a  four-year  course  of  theoretical 
preparation. 

In  the  opinion  of  the  author  this  would  be  highly  desirable,  because  details  and 
shop  practice  can  best  be  learned  in  the  shop,  where  experience  is  the  teacher.  It 
devolves  on  the  college  to  give  to  its  students  that  thorough  theoretical  training  which 
they  cannot  readily  acquire  in  practice,  a  feature  which  must  ultimately  distinguish 
the  college-bred  engineer  from  the  purely  practical  man. 

The  present  work,  which  is  the  gradual  evolution  of  years  of  labor,  thus  represents 
an  effort  to  place  before  the  profession  a  treatise  on  the  analysis  of  engineering  struc- 
tures which  is  based  on  the  most  advanced  theories  and  researches  of  the  present  time. 

To  enhance  the  educational  value  of  the  book,  each  subject  is  prefaced  by  a  few 
brief  historical  remarks  and  all  important  theorems  bear  the  names  of  their  originators, 
a  practice  which  has  been  shamefully  neglected  by  many  modern  writers. 

The  author  has  made  free  use  of  any  and  all  literature  bearing  on  the  various  sub- 
jects treated  and  hereby  expresses  his  grateful  acknowledgment  to  all  who  have  contrib- 
uted to  the  sum  total  of  our  present  knowledge.  Prominent  among  these  should  be 


vi  PREFACE 

mentioned  Hooke,  Bernoulli,  Lagrange,  Coulomb,  Navier,  Lame,  de  Saint- Venant, 
Clapeyron,  and  Menabrea,  as  the  founders  of  the  theories  of  elasticity  and  virtual  work. 

To  Clerk  Maxwell,  Castigliano,  Mohr,  Fraenkel,  Culmann,  Winkler,  Mueller-Breslau, 
Engesser,  Weyrauch,  Manderla,  Asimont,  Landsberg,  Melan,  Zimmerman,  Ritter,  Land, 
and  Mehrtens,  we  are  indebted  for  advancing  these  theories  to  their  present  state  of  per- 
fection and  usefulness.  Professor  Otto  Mohr  enjoys  the  high  distinction  of  being  the 
foremost  originator  of  novel  principles  in  the  analysis  of  engineering  structures. 

A  bibliography  of  the  works  used  by  the  author  will  be  found  at  the  end  of  this 
volume. 

The  author  has  avoided  a  sharp  differentiation  between  analytic  and  graphic  methods 
and  uses  both  without  discrimination,  aiming  always  to  secure  the  most  practical  solu- 
tion for  any  problem  in  hand. 

While  this  work  was  produced  with  very  great  care  and  diligence,  hoping  to  eliminate 
typographical  and  other  errors,  it  is  certain  that  some  have  escaped  detection.  The 
author,  therefore,  invites  the  kind  indulgence  of  the  reader  and  will  be  truly  thankful 
for  having  his  or  the  publishers'  attention  called  to  any  errors  that  may  be  discovered. 

Finally,  as  the  love  liberated  in  our  work  is  a  true  manifestation  of  character,  so 
may  the  present  production  reflect  the  character  of,  the  man,  the  engineer,  the  author. 

ITHACA,  N.  Y.,  September  9,  1910. 


DEFINITIONS  OF  TERMS  USED  THROUGHOUT  THIS  WORK 

FREQUENTLY  some  of  the  letters  are  employed  in  a  special  connection  other  than  here 
defined,  but  in  such  cases  the   definitions   locally  given  will  prevent  misunderstandings. 

ART.  3.    p  =  number  of  pin  points  of  any  frame. 
m  =  number  of  members  of  any  frame. 
n  =  number  of  redundant  conditions  in  any  structure. 
n'  =  number  of  external  redundants  in  any  structure. 
n"  =  number  of  internal  redundants  in  any  structure. 

e  =  number  of  elements  or  simple  frames  in  a  structure. 
Sr  =  number  of  necessary  reaction  conditions  for  any  structure. 
2  =  the  summation  sign. 
ART.  4.    S  =  total  stress  in  any  member  of  a  frame  resulting  from  any  loads  P  or  other 

causes  and  +  for  tension. 

&  =  the  stress  produced  in  any  member  by  applied  loads  P. 
/=  the  unit  stress  in  any  member,  +  for  tension. 

Z  =  length  of  any  member  or  span  of  a  structure  for  the  condition  of  no  stress. 
JZ  =  change  in  length  of  a  member  due  to  any  cause,  positive  for  elongation. 
F •—  cross-section  of  any  member,  prismatic  in  form. 
t=&  uniform  change  in  temperature  in  degrees,  +  for  rise. 
£  =  coefficient  of  expansion  per  degree  temperature. 
E=  Young's  modulus  of  elasticity  for  direct  stress. 
a  =  the  angle  which  a  member  makes  with  the  x  axis  of  coordinates. 
/?  =  the  angle  which  a  member  makes  with  the  y  axis  of  coordinates. 
m  =  any  point  of  a  structure  chosen  for  illustration. 
p  =  l/EF  =  the  extensibility  of  a  member  employed  for  brevity. 
.R  =  any  reaction  force,  or  condition,  of  a  structure  resulting  from  loads  P. 
.ft  =  any  reaction  force,  or  condition,  of  a  structure  resulting  from  loads  P. 
Jr  =  any  displacement  of  the  point  of  application  of  a  force  R  in  the  direction 

of  this  force. 

P  =  any  external  load  or  force  applied  to  a  structure. 
P  =  a  load  or  force  identical  in  position  and  point  of  application  with  the  force 

P  but  having  a  different  intensity. 
ART.  5.    J  =  any  displacement  of  a  pin  point. 

<?  =  the  displacement  of  the  point  of  application  of  any  force  P  in  the  direction 
of  this  force. 


viii         DEFINITIONS   OF  TERMS   USED   THROUGHOUT  THIS   WORK 

A  =  actual  work  =  force  times  distance  through  which  the  force  moves. 
Ae  =  ihe  positive  or  externally  applied  work  of  deformation  of  any  structure. 
.At-  =  the  negative  or  internally  overcome  work  of  deformation  of  any  structure. 
W  =  a  virtual  work  of  deformation. 

W  —  a  virtual  work  of  deformation  produced  by  a  conventional  case  of  loading. 
ART.  7.  Xa,  Xb,  Xc,  etc.,  are  the  stresses  in  redundant  members  or  reactions  A,  B,  C,  etc. 
£<,  =  the  stress  in  any  member  of  a  principal  system,  due  entirely  to  loads  P 
when  all  redundant  conditions  X  are  removed.      This  is  known  as  con- 
dition X  =  0. 

Sa  =  ihe  stress  in  any  member  of  a  principal  system,  due  entirely  to  the  conven- 
tional loading  Xa=l,  applied  to  the  principal  system.     Condition  Xn=l. 
Sb.  Sc,  etc.,  are  similarly  defined  for  conventional  loadings  Xb=l;  Xc=l,  etc. 
St  =  the  stress  in  any  member  of  a  principal  system  produced  by  a  uniform  change 

in  temperature. 

Rt  =  a  reaction  produced  by  a  uniform  change  in  temperature. 
Mt  =  &  moment  produced  by  a  uniform  change  in  temperature. 
R0,  Ra,  Rb,  etc.,  have  definitions  analogous  to  those  given  for  S0,  Sa,  Sb,  etc. 
M0f  Ma,  Mb,  etc.,  are  moments  similarly  defined  for  the  conventional  loadii^s. 
da,  $b,  $c  are  changes  in  the  lengths  of  redundant  members  or  supports  Xa,  Xbf 

Xe,  respectively. 

L  represents  the  length  of  span  of  a  composite  structure. 

ART.  8.  dma  =  the  displacement  of  the  point  of  application  m  of  any  load  Pm,  in  the 
direction  of  this  load,  when  the  principal  system  is  loaded  only  with 
Xa  =  1 .     Condition  Xa  =  1 . 
Smb  =  a  similar  displacement  of  the  point  m  for  the  conventional  loading  Xb  =  1 . 

Condition  Xb=l. 

<5TOC  =  the  same  for  condition  Xc=  1. 

daa  =  the  change  in  length  of  the  member  a  for  condition  Xa—l. 
(50j>  =  the  change  in  length  of  the  member  a  for  condition  Xb=l. 
d&a  =  the  change  in  length  of  the  member  b  for  condition  Xa=l,  etc. 
dat  =  the  change  in  length  of  the  member  a  resulting  from  a  change  in  temperature 
of  t°,  when  the  principal  system  is  not  otherwise  loaded,  hence  X  =  0,  P  =  0. 
dbt,  $ct  are  similarly  defined  for  members  b  and  c  respectively, 
ART.  13.  fx,  fy,  /2  =  the  unit  normal  stresses  in  the  directions  X,  Y,  and  Z,  respectively , 

for  any  isotropic  body. 

TXV  and  rx:s  =  the  unit  tangential  stresses  in  the  YZ  plane. 
rvx  and  rvz  =  the  unit  tangential  stresses  in  the  XZ  plane. 
TZX  and  Tzv  =  the  unit  tangential  stresses  in  the  XY  plane. 
G  =  the  modulus  of  tangential  stress. 

m  =  the  Poisson  number  or  ratio  of  lateral  to  longitudinal  deformation. 
ART.  14.  N  =  &  normal  thrust  on  any  section. 
R  =  any  oblique  thrust. 
Q  =  a  tangential  force  or  shear. 
I  =  the  moment  of  inertia  of  anv  cross-section. 


DEFINITIONS   OF  TERMS  USED   THROUGHOUT   THIS   WORK  ix 

t/  =  an  ordinate  usually  the  distance  from  the  neutral  axis  to  the  extreme  fiber 

of  a  cross-section. 
h  =  height  of  a  section. 
6  =  base  or  breadth  of  a  section. 

/?  =  coefficient  of  shearing  strain,  Eq.  (ML)  and  Eq.  (14M). 
ART.  16.  w  =  weight  of  a  moving  body. 

H  =  height  of  fall.     Later  used  to  represent  the  horizontal  thrust  of  an  arch, 

also,  the  pole  distance  of  a  force  polygon. 
v  =  velocity  at  instant  of  impact. 
g  =  acceleration  due  to  gravity. 
ART.  17.  jj  =  any  ordinate  to  an  influence  line. 
d  =  a  panel  length. 

p  =  uniform  moving  load  per  unit  of  length. 
ART.  18.  m  =  a  load  point,  being  any  one  of  the  many  possible  points  of  application  of 

a  certain  moving  load. 
n  =  the  point  for  which  art  influence  line  is  drawn.     Also,  the  location  of  a  section 

under  consideration. 
ART.  20.   i  =  the  load  divide  for  positive  and  negative  effects.     For    arches    the    load 

divide  is  sometimes  designated  by  d. 
ART.  21.   r  =  the  lever  arm  of  a  member  in  the  equation  S  =  M/r. 

Sa  =  ihe  stress  in  any  member  S  due  to  a  reaction  unity  at  A,  when  the  load 

producing  this  reaction  is  to  the  right  of  the  section. 
Si,  =  the  stress  in  the  same  member  due  to  a  reaction  unity  at  B,  when  the  load 

producing  this  reaction  is  to  the  left  of  the  section. 

ART.  28.  e  =  the  kernel  point  for  the  extreme  fiber  of  the  extrados  of  a  solid  web  arch  section. 

i  =  the  kernel  point  for  the  extreme  fiber  of  the  intrados  of  the  same  arch  section. 

ART.  35.  w  =  an  elastic  load,  representing  an  abstract  number  without  any  particular  unit 

of  measure. 

ART.  45.  <S0 = the  stress  in  any  member  of  a  statically  indeterminate  frame  for  the  con- 
dition J£  =  0  and  P=l. 
M0=  a  moment  for  the  same  conditions. 
A0=  the  reaction  at  A  for  the  same  conditions. 

/x  =  an  influence  line  factor  by.  which  all  ordinates  must  be  multiplied  to  obtain 
the  actual  influences. 


CONTENTS 


PAGE 

PREFACE v 

DEFINITIONS  OF  TERMS vii 

INTRODUCTION 1 

CHAPTER   I 
REACTIONS  AND  REDUNDANT  CONDITIONS 

1.  Reaction  Conditions 5 

2.  Structural  Redundancy 6 

3.  Tests  for  Structural  Redundancy 7 

CHAPTER  II 
THEOREMS,  LAWS,  AND  FORMULAE  FOR  FRAMED  STRUCTURES 

4.  Elastic  Deformations,  Fundamental  Equations 11 

5.  General  Work  Equations  for  any  Frame 14 

6.  Displacements  of  Points.     Statically  Determinate  Structures 18 

7.  Indeterminate  Structures,  by  Mohr's  Work  Equation 19 

8.  Indeterminate  Structures,  by  Maxwell's  Law 24 

9.  Prof.  Maxwell  s  Theorem  (1864) 27 

10.  Theorems  Relating  to  Work  of  Deformation 29 

(a)  Menabrea's  law  (1858).     Theorem  of  least  work 29 

(b)  Castigliano's  law  (1879).     Derivative  of  the  work  equation 30 

11.  Temperature  Stresses  follow  Menabrea's  and  Castigliano's  Laws 31 

12.  Stresses  Due  to  Abutment  Displacements 32 

CHAPTER  III 

THEOREMS,  LAWS,  AND  FORMULAE  FOR  ISOTROPIC  BODIES 

1 3-  General  Work  Equations 34 

14.  Work  of  Deformation  due  to  Shearing  Stress 38 

15.  Work  of  Deformation  for  any  Indeterminate  Straight  Beam 41 

16.  Work  of  Deformation  due  to  Dynamic  Impact 43 

xi 


xii  CONTENTS 

CHAPTER  IV 
INFLUENCE  LINES  AND  AREAS  FOR  STATICALLY  DETERMINATE  STRUCTURES 

PAGE 

1 7.  Introductory 46 

18.  Influence  Lines  for  Direct  Loading 4<S 

19.  Influence  Lines  for  Indirect  Loading 50 

20.  The  Load  Divide  for  a  Truss 51 

21.  Stress  Influence  Lines  for  Truss  Members    52 

22.  Reaction  Summation  Influence  Lines 54 

23.  Positions  of  a  Moving  Train  for  Maximum  and  Minimum  Moments  57 

24.  Positions  of  a  Moving  Train  for  Maximum  and  Minimum  Shears 62 

CHAPTER  V 
SPECIAL  APPLICATIONS  OF  INFLUENCE  LINES  TO  STATICALLY  DETERMINATE  STRUCTURES 

25.  Double  Intersection  Trusses 65 

26.  Cantilever   Bridges 67 

27.  Three-hinged  Framed  Arches 72 

28.  Three-hinged,  Solid  Web  and  Masonry  Arches 78 

29.  Skew  Plate  Girder  for  Curved  Double  Track 80 

30.  Trusses  with  Sub-divided  Panels S3 

CHAPTER  VI 
DISTORTION  OF  A  STATICALLY  DETERMINATE  FRAME  BY  GRAPHICS 

31 .  Introductory 87 

32.  Distortions  due  to  Changes  in  the  Lengths  of  the  Members 87 

33.  Rotation  of  a  Rigid  Frame  about  a  Fixed  Point 89 

34.  Special  Applications  of  Williot-Mohr  Diagrams 91 

CHAPTER  VII 
DEFLECTION  POLYGONS  OF  STATICALLY  DETERMINATE  STRUCTURES  BY  ANALYTICS  AND  GRAPHICS 

35.  Introductory 9!) 

36.  First  Method,  according  to  Prof.  Mohr 100 

(a)  Influence  on  the  deflections  due  to  chord  members 100 

(6)   Influence  on  the  deflections  due  to  web  members .  102 

(c)   The  deflection  polygon  for  the  loaded  chord 104 

37.  Second  Method,  according  to  Prof.  Land , 107 

(a)  The  w  loads  in  terms  of  the  unit  stresses  in  the  members  .  .   107 

(6)  To  find  the  changes  in  the  angles  of  a  triangle 1 07 

(c)   To  evaluate  the  elastic  loads  w  in  terms  of  the  angle  changes 109 

38.  Horizontal  Displacements • 114 

39.  Deflections  of  Solid  Web  Beams  ..                                                                                                .  117 


CONTENTS  xiii 

CHAPTER  VIII 
DEFLECTION  INFLUENCE  LINES  FOR  STATICALLY  DETERMINATE  STRUCTURES 

PAGE 

40.  Influence  Lines  for  Elastic  Displacements 122 

41.  Special  Applications  of  Displacement  Influence  Lines 124 

(a)  Cantilever  beam,  deflection  polygon 124 

(b)  Simple  truss,  displacement  of  any  pin  point  in  any  direction 125 

(c)  Three-hinged  arch,  rotation  between  two  lines 125 

CHAPTER  IX 
INFLUENCE  LINES  FOR  STATICALLY  INDETERMINATE  STRUCTURES 

42.  Influence  Lines  for  One-  Redundant  Condition 127 

(a)  When  the  redundant  condition  is  internal 128 

(b)  When  the  redundant  condition  is  external 128 

43.  Influence  Lines  for  Two  Redundant  Conditions 130 

44.  Simplification  of  Influence  Lines  for  Multiple  Redundancy 132 

(a)  Structures  having  two  redundant  conditions 133 

(b)  Structures  having  three  redundant  conditions 135 

45.  Stress  Influence  Lines  for  Structures  Involving  Redundancy 137 

(a)  Multiple  redundancy 137 

(6)  One  redundant  condition 139 

CHAPTER  X 
SPECIAL  APPLICATIONS  OF  INFLUENCE  LINES  TO  STATICALLY  INDETERMINATE  STRUCTURES 

46.  Simple  Beam  with  One  End  Fixed  and  Other  End  Supported 140 

47.  Plate  Girder  on  Three  Supports 143 

48.  Truss  on  Three  Supports 148 

49.  Two-hinged  Solid  Web  Arch  or  Arched  Rib 153 

50.  Two-hinged  Spandrel  Braced  Arch 166 

51.  Two-hinged  Arch  with  Cantilever  Side  Spans 176 

52.  Fixed  Framed  Arches 178 

CHAFFER  XI 
DESIGN  OF  STATICALLY  INDETERMINATE  STRUCTURES 

53.  Methods  for  Preliminary  Designing 194 

54.  On  the  Choice  of  the  Redundant  Conditions 202 

55.  Solution  of  the  General  Case  of  Redundancy 203 

56.  Effect  of  Temperature  on  Indeterminate  Structures 206 

57.  Effect  of  Shop  Lengths  and  Abutment  Displacements 208 


xiv  CONTENTS 


CHAPTER  XII 

STRESSES  IN  STATICALLY  DETERMINATE  STRUCTURES 

PAGE 

58.  Dead  Load  Stresses 210 

(a)  General  considerations 210 

(6)   Aug.  Hitter's  methods  of  moments  (1860) 211 

(c)   The  method  of  stress  diagrams 213 

59.  Live  Load  Stresses 218 

(a)  The  critical  positions  of  a  train  of  moving  loads 218 

(b)  The  method  of  influence  lines 218 

(c)  Discussion  of  methods  in  common  use 219 

(d)  The  author's  method 210 


7  CHAPTER  XIII 

SECONDARY  STRESSES 

60.  The  Nature  of  Secondary  Stresses —-.-. . .  226 

61.  Secondary  Stresses  in  the  Plane  of  a  Truss  due  to  Riveted  Connections 227 

62.  Secondary  Stresses  in  Riveted  Cross  Frames  of  Trusses 244 

(a)  Dead  and  live  load  effects 244 

(6)  Wind  effects 247 

63.  Secondary  Stresses  due  to  Various  Causes 251 

(a)  Bending  stresses  in  the  members  due  to  their  own  weight 251 

(6)   Secondary  stresses  in  pin-connected  structures 253 

(c)  The  effect  of  eccentric  connections 255 

(d)  Effect  of  temperature  and  erroneous  shop  lengths 256 

64.  Additional  Stresses  due  to  Dynamic  Influences 256 

(a)  Forces  acting  longitudinally  on  a  structure 256 

(6)  Forces  acting  transversely  to  a  structure 258 

(c)  Forces  acting  vertically  on  a  structure 262 

(d)  Fatigue  of  the  material 265 

(e)  Unusual  load  effects 266 

65.  Concluding  Remarks 207 

(a)  Features  in  design  intended  to  diminish  secondary  and  additional  stresses 267 

(b)  Final  combination  of  stresses  as  a  basis  for  designing 268 


CHAPTER  XIV 
MITERING  LOCK  GATES 

66.  The  Nature  of  the  Problem 271 

67.  The  Theory  of  the  Analysis 273 

68.  Example :  Upper  Gate,  Erie  Canal  Lock  No.  22 279 


CONTENTS  xv 

CHAPTER  XV 

FIXED  MASONRY  ARCHES 

PAGE 

69.  General  Considerations 298 

70.  Modern  Methods  of  Construction 301 

71.  Preliminary  Designs 304 

72.  Determination  of  the  Redundant  Conditions  by  the  Theory  of  Elasticity 309 

73.  Co-ordinate  Axes  and  Elastic  Loads 314 

74.  Influence  Lines  for  Xa,  Xb,  Xc,  and  Mm 317 

75.  Temperature  Stresses 318 

76.  Stresses  on  any  Normal  Arch  Section 320 

77.  Maximum  Stresses  and  Critical  Sections 322 

78.  Resultant  Polygons 324 

79.  Example :  150-foot  Concrete  Arch 326 

(a)  Given  data  arid  preliminary  design 326 

(6)  Analysis  of  the  preliminary  design  by  the  theory  of  elasticity,  treating  the  structure 

as  symmetric 331 

(c)   Analysis  as  under  (fr),  treating  the  structure  as  unsymmetric 346 

BIBLIOGRAPHY 

Theory  of  Elasticity 359 

Graphostatics 359 

General  Treatises .' , 360 

Secondary  and  Additional  Stresses 360 

Special  Treatises  on  Arches 361 


KINETIC   THEORY 


OF 


ENGINEERING    STRUCTURES 


INTRODUCTION 

The  work  of  deformation  constitutes  the  basis  for  the  solution  of  all  those  problems 
which  involve  the  elastic  properties  of  the  material  and  which  are  not  susceptible  to 
analysis  by  the  methods  of  pure  statics. 

Work  is  the  product  of  a  constant  force  into  the  displacement  of  its  point  of  applica- 
tion in  the  direction  of  the  force. 

When  a  prismatic  body  of  length  I  and  cross-section  F  is  either  lengthened  or 
shortened  a  small  amount  M  within  the  elastic  limit,  by  a  force  S  gradually  produced 
without  impact,  then  a  certain  amount  of  work  or  kinetic  energy  is  stored  up  in  the 
body.  In  the  instant  when  the  force  ceases  to  alter  the  length  a  condition  of  static 
equilibrium  is  established  and  the  total  kinetic  energy  stored  then  assumes  the  form 
of  potential  energy  or  energy  of  position.  WThen  the  force  is  gradually  released  and 
the  body  is  allowed  to  resume  its  original  length,  the  potential  energy  is  liberated. 

The  work  of  deformation  or  applied  kinetic  energy  may  be  thus  expressed: 


The  principle  of  virtual  work.  When  a  rigid  body  is  set  in  motion  by  any  group 
of  forces,  then  the  kinetic  energy  imparted  is  equal  to  the  algebraic  sum  of  all  the  work 
performed  by  the  several  applied  forces. 

A  real  displacement  can  never  result  in  zero  work,'  because  motion  cannot  exist 
without  the  presence  of  a  force,  and  the  combination  of  force  and  motion  constitutes 
work. 

However,  for  a  state  of  equilibrium  wherein  the  applied  forces  do  not  actually  alter 
the  motion,  a  certain  equilibrium  proposition  may  be  formulated  by  considering  possible 
displacements  that  might  be  produced  but  may  or  may  not  be  real.  Such  displacements 
are  called  virtual,  and  any  work  resulting  therefrom  is  called  virtual  work. 


2  KINETIC    THEORY   OF    ENGINEERING   STRUCTURES 

The  fundamental  principle  of  statics  may,  therefore,  be  stated  thus:  A  body  at 
rest  remains  at  rest  when  for  any  infinitesimally  small  but  possible  displacement  the 
summation  of  the  work  of  the  applied  forces  is  zero. 

Letting  P  represent  any  force  and  d  any  possible  virtual  displacement  of  its  point 
of  application  in  the  direction  of  the  force,  then,  according  to  Lagrange  (1788),  the  law 
of  virtual  work  is  represented  by  the  following  equation: 

0  .............     (2) 


The  law    of  virtual  velocities  follows  from  the  law  of  virtual  work  by  introducing  the 
element  of  time,  giving 

0  ........     .....     (3) 


The  relations  thus  existing  between  the  possible  conditions  of  equilibrium  of  a  rigid 
body  and  the  virtual  displacements  resulting  from  any  cause,  lead  to  important  geometric 
relations  which  are  generally  applicable  to  all  branches  of  technical  mechanics. 

Clapeyron's  law  (1833),  which  covers  the  entire  work  of  the  external  forces  and 
internal  stresses  acting  on  any  frame  or  body  in  equilibrium,  is  derived,  from  Eq.  (1), 
by  summing  all  the  partial  effects  for  an  entire  structure.  Then,  by  the  "  doctrine  of  the 
conservation  of  energy,"  the  applied  work  Ae  must  equal  A;  the  work  overcome,  giving 

.........     (4) 


wherein  the  loads  P,  stresses  S,  displacements  d  projected  onto  the  directions  of  the 
loads,  and  the  changes  Al  in  the  lengths  of  the  members,  all  result  from  a  simultaneous 
condition  of  loading  for  a  case  in  equilibrium. 

Proceeding  from  Clapeyron's  law.  Professor  Clerk  Maxwell,  in  1864,  derives  his 
important  law  of  reciprocal  displacements  and  Professor  Otto  Mohr,  in  1874,  evolved  his 
two  work  equations  which  together  with  Maxwell's  law,  afford  a  complete  solution  for 
all  problems  relating  to  frames  and  isotropic  bodies. 

Chapter  I  deals  with  the  general  relations  existing  between  a  frame  or  beam,  arid 
its  supports,  furnishing  the  definitions  and  necessary  criteria  for  distinguishing  between 
statically  determinate  and  statically  indeterminate  structures  or  those  involving  redundant 
members  or  redundant  reaction  conditions. 

Chapter  II  then  presents  the  derivation  and  discussion  of  all  the  general  laws  per- 
taining to  framed  structures  and  these  are  applied  to  isotropic  bodies  in  Chapter  III. 

Chapters  IV,  V,  and  VIII  are  devoted  to  influence  lines  for  determinate  structures, 
while  Chapters  VI  and  VII  treat  of  the  various  problems  relating  to  distortions  and 
deflection  polygons  of  frames  and  solid  web  structures. 

Chapters  IX  and  X  apply  the  various  theories  and  methods  to  the  analysis  of 
statically  indeterminate  structures  chiefly  by  the  use  of  influence  lines. 

In  Chapter  XI,  the  general  question  of  designing  a  statically  indeterminate  structure 
is  then  taken  up  and  Chapter  XII  deals  with  stresses  in  statically  determinate  structures, 


INTRODUCTION  3 

taking  advantage  of  some  of  the  more  advanced  ideas  presented  in  the  earlier  chapters. 
Special  attention  is  directed  to  Art.  59,  dealing  with  stresses  due  to  concentrated  live 
loads. 

Chapter  XIII  gives  a  brief  treatment  of  secondary  stresses  and  the  two  last  chapters 
are  devoted  to  the  problems  of  mitering  lock  gates  and  fixed  masonry  arches. 

It  is  more  than  likely  that  the  use  of  the  word  kinetic  in  the  title  of  the  present 
volume  will  be  questioned  by  some.  However,  after  long  and  careful  consideration 
the  author  concluded  that  he  was  justified  in  employing  the  term  without  really  depart- 
ing from  the  best  modern  dictionary  definitions.  It  is  true  that  in  hydraulics  the  term 
kinetic  energy  is  exclusively  confined  to  represent  the  static  equivalent  h=v2/2g  of  a 
dynamic  head.  In  mechanics  generally,  the  expression  is  almost  limited  to  the  term 
Wv2/2g=mv2/2,  representing  the  potential  or  kinetic  energy  of  amoving  mass.  Other- 
wise the  term  is  not  met  with. 

While  the  author  employs  the  term  kinetic  in  connection  with  engineering  structures 
which  are,  generally  considered,  at  rest,  it  is  precisely  with  the  same  idea  above  expressed 
in  the  mechanical  sense.  That  is,  a  bridge  or  other  structure  composed  of  elastic  mate- 
rial ceases  to  be  at  rest  the  moment  it  is  acted  upon  by  external  forces.  The  distinction 
between  the  previous  and  present  uses  of  the  term  is,  therefore,  only  relative,  being 
here  applied  in  a  different  field  and  to  a  special  class  of  bodies,  where  the  displacements 
are  small  and  follow  certain  empiric  laws  depending  on  the  elastic  properties  of  the 
material.  Strictly  speaking,  then,  we  have  no  license  to  regard  an  elastic  structure  as 
statically  at  rest  only  when  it  is  neither  stressed  nor  distorted. 

The  equation  of  work  is  thus  applied  to  a  structure  in  the  sense  that  the  latter  is 
a  mechanical  contrivance  or  machine  in  motion,  instead  of  an  inelastic  body  at  rest. 
It  is  true  that  this  motion  prevails  only  while  a  change  in  the  static  balance  is  taking 
place,  as  when  loads  are  added  to,  or  removed  from,  the  structure,  but  this  may  apply  to 
any  mechanism  of  interrupted  activity. 

Therefore,  while  we  speak  of  a  structure  as  in  static  equilibrium, 'we  may  also  speak 
of  it  as  in  a  state  of  dynamic  equilibrium,  a  state  which  the  structure  assumes  in  the  instant 
that  the  superimposed  loads  do  not  produce  any  further  elastic  deformations.  The 
same  would  be  true  when  all  the  loads  are  entirely  removed,  in  which  case  the  dynamic 
equilibrium  returns  to  the  special  state  of  static  equilibrium.  This  is  merely  a  broader 
viewpoint  of  the  subject,  embracing  at  once  all  of  the  conditions  as  they  really  exist 
in  a  frame  sustaining  loads.  The  term  kinetic  thus  appears  quite  appropriate  to  the 
use  and  application  here  made. 

For  the  same  structure  the  magnitude  of  the  deformation  of  the  deflection  is  a 
direct  function  of  the  loads,  the  unit  stresses  which  they  produce,  and  the  elasticity  of 
the  material. 

On  the  other  hand,  the  stiffness  of  a  given  structure  is  independent  of  the  magnitudes 
of  the  unit  stresses,  but  depends  entirely  on  the  ability  of  the  structure  to  resist  stress 
and  this  in  turn  is  a  function  of  the  elasticity  of  the  material  and  the  geometric  shape 
of  the  structure. 

Theoretically  speaking,  and  disregarding  economy,  a  structure  of  maximum  stiff- 
ness is  one  of  infinite  height  compared  to  its  span,  resulting  in  zero  stress;  and  con- 


4  KINETIC   THEORY   OF   ENGINEERING   STRUCTURES 

versely,  a  structure  of  minimum  stiffness  is  one  of  zero  height  involving  infinite 
stress. 

Maximum  stress  never  occurs  simultaneously  in  the  chords  and  web  of  any  structure. 
The  maximum  total  load  produces  maximum  chord  stresses,  but  only  a  small  fraction 
of  the  maximum  web  stress.  Hence,  the  influence  of  the  web  on  the  total  deflection 
of  a  bridge  is  always  small  as  compared  with  that  produced  by  the  chords  and  is  fre- 
quently negligible. 

Throughout  the  present  treatise,  exepting  Chapter  XIII,  all  members  of  a  frame 
are  supposedly  connected  at  their  ends  by  frictionless  pins,  and  no  account  is  taken  of 
the  so-called  secondary  stresses  occasioned  in  the  members  by  friction  on  these  pins  or 
by  the  rigidity  of  riveted  connections.  Chapter  XIII  deals  separately  with  the  secondary 
stresses  induced  by  riveted  connections  and  friction  on  pins,  etc. 

Unless  otherwise  specifically  stated,  a  clockwise  moment  of  forces  on  the  left  of  a 
section  is  regarded  as  positive,  so  also  an  upward  shear  on  the  left  side  of  a  section. 
Tensile  stress  and  elongation  are  positive,  while  compressive  stress  and  contraction  are 
negative.  Work  is  positive  when  the  direction  of  a  force  coincides  with  the  direction 
of  the  displacement  of  its  point  of  application,  independently  of  the  algebraic  signs  of 
the  forces  or  stresses  and  the  displacements. 

The  nomenclature  is  uniform  throughout,  though  in  a  few  instances  slight  departures 
from  American  custom  became  necessary  owing  to  conflicts  encountered  in  combining 
so  many  subjects  in  one  book.  The  definitions  of  terms  are  always  given  in  appropriate 
places  and  also  in  summary  form  following  the  table  of  contents. 

Equations  and  figures  are  designated  by  letters  and  numbers  facilitating  the  matter 
of  locating  cross-references.  The  number  refers  to  the  article  and  the  letter  to  the 
position  of  the  equation,  figure  or  table  in  that  article.  The  article  number  appears  in 
the  heading  of  every  page. 


ART.  1.     REACTION   CONDITIONS 

The  general  purpose  of  engineering  structures,  of  the  class  here  considered,  is  to 
carry  loads  over  otherwise  impassable  distances  to  certain  fixed  points. 

The  distances  are  called  the  spans,  and  the  fixed  points,  to  which  the  loads  are 
finally  transmitted,  are  called  the  supports  or  reactions. 

It  is  thus  clear  that  the  sum  total  of  the  superimposed  loads,  together  with  the 
weight  of  the  structure,  must  exactly  equal  the  sum  of  the  vertical  reaction  forces  when 
equilibrium  exists. 

The  nature  and  purpose  of  a  structure  determine  the  manner  in  which  it  must  be 
supported  to  insure  stability  under  all  circumstances  that  are  likely  to  occur. 

This  can  always  be  accomplished  by  one  of  three  types  of  supports  known,  respec- 
tively, as  movable,  hinged,  and  fixed.  They  are  illustrated  in  Figs.  IA,  IB,  and  Ic. 


FIG.  IA. 


FIG.  IB. 


FIG.  Ic. 


Fig.  IA  shows  the  type  of  support  known  as  movable,  which  can  exert  a  single 
upward  reaction  only.  Any  lateral  force  applied  to  such  a  support  would  result  in 
horizontal  displacement  provided  the  roller  is  frictionless. 

Fig.  IB  shows  a  hinged  support  and  can  resist  horizontal  and  vertical  forces  or 
any  resultant  of  these. 

Fig.  Ic  represents  a  fixed  support  which  may  be  made  to  resist  any  direct  forces 
and  a  bending  moment. 

Statically  considered,  the  first  type  satisfies  one  stability  condition,  the  second 
type  offers  two  such  conditions,  and  the  third  type  supplies  three  stability  conditions. 
These  conditions  may  be  represented  by  forces  as  shown  in  Figs.  ID,  IE,  and  IF. 

Let  r  represent  a  single  vertical  or  horizontal  force  to  be  known  as  a  reaction  con- 
dition. Then  the  three  types  of  supports  will  involve,  respectively,  one,  two,  and  three 
such  conditions.  Each  reaction  condition  may  be  represented  by  a  short  link,  or  member, 
which  may  transmit  direct  stress  only;  that  is  pure  tension  or  compression. 

5 


KINETIC  THEORY  OF    ENGINEERING   STRUCTURES 


CHAP.  I 


Hence,  for  a  single  truss  on  two  supports,  with  one  end  movable  and  the  other 
end  hinged,  the  number  of  reaction  conditions  is  Sr=3.  Also,  for  a  two-hinged  arch, 
with  both  ends  on  hinged  supports,  2r=4;  and  for  an  arch  without  hinged  ends,  com- 
monly called  a  fixed  arch,  2r=6. 


FIG.  ID. 


FIG.  IE. 


r 
FIG.  IF. 


ART.  2.  STRUCTURAL  REDUNDANCY 

Any  structure,  statically  considered,  may  be  classed  as  determinate  or  indeterminate, 
depending  on  the  manner  in  which  it  is  supported,  and  the  arrangement  of  the  members 
employed  to  form  the  structure. 

A  structure  is  externally  indeterminate  when  it  involves  more  than  three  reaction 
conditions  for  a  single  truss  element  or  frame.  This  implies  that  the  reactions  are 
counted  with  the  external  forces. 

A  structure  is  internally  indeterminate  when  it  includes  more  members  within  its 
frame  than  are  required  for  internal  stability  according  to  the  conditions  for  static  equilib- 
rium, regardless  of  the  reactions. 

Hence  external  and  internal  redundancy  may  exist  simultaneously.  The  terms 
redundant  and  indeterminate,  as  here  used,  are  synonymous  with  impossible,  only  in  so 
far  as  the  laws  of  pure  statics  are  concerned.  However,  the  modifying  term  static  may 
sometimes  be  omitted  for  the  sake  of  brevity.  Thus  indeterminate  will  always  mean 
statically  indeterminate. 

The  conditions  for  static  equilibrium  applied  to  any  structure  as  a  whole  are  repre- 
sented by  the  following  equations  : 


27=0,   and   2M=0,  ......  (2A) 


wherein  2#  is  the  sum  of  all  the  horizontal  components  of  the  external  forces,  includ- 
ing the  reactions;  2F  is  the  sum  of  all  the  vertical  components  of  these  forces  and  reac- 
tions; and  SM  is  the  sum  of  the  moments  of  these  forces  and  reactions  about  any  point 
in  the  plane  of  the  structure.  The  sums  are  always  algebraic  sums. 

These  condition  equations  apply  equally  to  all  forces  (externally  applied  and  internal 
stresses)  active  about  any  pin  point  of  any  structure. 

When  the  external  forces  are  not  thus  balanced  among  themselves  and  with  the 
internal  stresses,  then  a  structure  is  said  to  be  statically  under  determinate  or  to  be  unstable. 

A  structure  is,  therefore,  externally  determinate  when  the  three  condition  equations 
for  static  equilibrium  suffice  to  determine  the  reactions. 


ART.  3  REACTIONS   AND  REDUNDANT   CONDITIONS  7 

A  structure  is  internally  determinate  when  its  members  are  arranged  to  form  triangles 
in  such  manner  that  the  successive  removal  of  pairs  of  members  uniting  in  a  point, 
finally  reduces  the  structure  to  a  single  triangle.  The  triangle  is  thus  seen  to  be  the 
primary  element  of  all  determinate  systems. 

A  structure  is  in  every  sense  determinate  when  all  reactions  and  stresses  can  be  expressed 
purely  as  functions  of  the  externally  applied  loads,  unaffected  by  all  temperature  changes 
or  temperature  inequalities,  and  small  reaction  displacements. 

ART.  3.  TESTS  FOR  STRUCTURAL  REDUNDANCY 

Let  p=  number  of  pin  points  of  any  given  frame. 
m=  number  of  members  in  the  frame. 
n=  total  number  of  redundant  conditions. 
ri  =  number  of  external  redundant  conditions. 
n"  =  number  of  internal  redundant  conditions. 

e=  number  of  elements  or  simple  frames  in  a  structure. 
2r=  number  of  necessary  reaction  conditions. 
Then  for  any  statically  determinate  structure 


which  condition  must  be  satisfied  in  any  event,  and  when  satisfied,  the  structure  is 
always  statically  determinate  but  not  always  stable.  This  exception  will  be  discussed 
below. 

When  2p>m  +  Zr,       ...........     (SB) 


then  the  structure  is  not  in  stable  equilibrium  and  may  be  called  statically  insufficient. 
When  2p<w  +  Sr  then  the  structure  involves  redundant  conditions  or  members. 
the  total  number  of  which  is  given  by  the  equation, 

n=m  +  Zr  —  2p,   ...........     (3c) 

wherein  a  negative  n  would  indicate  static  insufficiency. 

The  number  of  necessary  reaction  conditions  r  for  any  structure  depends  on  the 
number  of  frame  elements  or  simple  frames  of  which  the  structure  may  be  composed. 
Hence  in  order  to  distinguish  whether  the  n  redundant  conditions  are  external,  internal, 
or  both,  it  will  be  necessary  to  determine  one  or  the  other  of  these  by  a  separate 
criterion.  This  criterion  is  readily  established  for  the  reactions.  Thus  in  Art.  1,  Sr=3 
was  found  to  express  the  number  of  required  reaction  conditions  for  a  simple  truss  on 
two  supports. 

A  composite  structure,  which  may  be  composed  of  any  number  e  of  simple  truss 
frames,  each  one  of  which  is  directly  supported  by  piers  at  one  or  two  points,  will  always 
have  more  than  three  reaction  conditions.  The  requisite  number  of  reaction  conditions 
for  any  statically  determinate  structure  is 

£r=3+e-l=e+2.  .     .  (3D) 


8 


KINETIC  THEORY   OF  ENGINEERING  STRUCTURES 


CHAP.  I 


The  number  of  external  redundant  conditions  n',  for  any  given  structure  of  e  elements, 
is  thus: 

w'  =  2r-e  — 2, (3E) 


and  hence  the  number  of  internal  redundant  conditions  must  be 


n    =n—n. 


OF) 


The  number  of  elements  entering  into  the  makeup  of  any  structure  is  thus  an  important 
consideration  in  establishing  whether  the  redundant  conditions,  indicated  by  Eq.  (3c), 
are  external  or  internal.  In  general,  a  truss  element  may  be  denned  as  a  frame  or  girder 
which  in  itself  constitutes  a  simple  truss  or  beam.  The  illustrations  given  below  will 
serve  to  amplify  this  definition. 

The  exception  to  Eq.  (3A),  when  a  structure  is  apparently  determinate  but  unstable, 
will  now  be  discussed. 


FIG.  SA. 


FIG.  3s. 


The  three-hinged  arch,  in  Fig.  SA,  is  in  every  sense  a  determinate,  stable  structure 
for  which  the  horizontal  thrust  H  =Pl/2f.  All  the  reactions  A,  B,  and  H  are  finite 
quantities  and  the  stresses  in  the  several  members  are  also  finite.  This  will  always 
remain  true  except  when  the  middle  ordinate  /=0,  whence  //  =  sc  . 

This  special  case  is  illustrated  in  Fig.  SB,  where  ACB  is  a  straight  line  and  the  load 
P  cannot  produce  stress  without  causing  a  slight  rotation  of  the  two  elements  I  and  II 
about  the  points  A  and  B.  Hence  equilibrium  does  not  exist  until  rotation  through 
a  small  arc  ds  has  taken  place  and  the  stress  along  AC  and  CB  is  then  infinite.  Struc- 
tures involving  this  peculiar  feature  of  the  well-known  "  toggle  joint  "  are  regarded  tem- 

porarily unstable,  and  equilibrium  in  them 
can  be  established  only  after  certain 
infinitesimally  small  displacements  have 
taken  place. 

Another  striking  example  of  this  kind 
is  shown  in  Fig.  3c,  where  the  supports  A 
and  C  are  hinged,  and  the  support  B  is 
movable,  and  hence  2r=5.  Also  w  =  19 
and  p  =  \2,  making  n=w  +  Hr  —  2p=0. 

This    structure  is  made  up    of    three 
elements  I,  II  and  III,  each  one  of  which 
may  be  regarded  as  a  simple  frame  in  direct  contact  with  a  support.     Hence  e=3  and 


This  proves  the  structure  to  be  internally  as  well  as  externally  statically  determinate, 


ART.  3 


REACTIONS  AND   REDUNDANT   CONDITIONS 


9 


though  a  condition  of  temporary  instability  exists  because  the  point  k  is  common  to 
the  paths  of  the  three  elements  in  the  first  instant  of  rotation  about  their  respective 
centers  of  rotation  at  A,  B,  and  C. 


FIG.  3o. 


FIG.  3o. 


FIG.  3K. 


FIG.  SQ 


FIG.  SE. 


FIG.  3F. 


FIG.  3n. 


FIG.  3j. 


FIG.  SL. 


FIG.  3n. 


The  point  k  may  thus  be  regarded  as  an  imaginary  crown  hinge  for  a  three-hinged 
arch  AkC.  Hence  a  load  P,  acting  at  D,  will  cause  a  small  rotation  of  the  three  elements 
before  equilibrium  is  established  and  then  the  stresses  become  infinite. 


10 


KINETIC   THEORY   OF   ENGINEERING  STRUCTURES 


CHAP.  I 


The  only  loads  which  will  not  cause  rotation  are  those  passing  through  k,  but  for 
any  of  these  loads,  there  will  result  an  infinite  number  of  conditions  of  equilibrium  and 
as  many  different  systems  of  stresses. 

The  following  illustrations  are  presented  for  the  purpose  of  showing  various  applica- 
tions of  Eqs.  (3c)  and  (SE).  Figs.  3D  to  3M  are  all  examples  of  simple  truss  elements 
for  which  e  =  l.  Fig.  SN  has  six  such  elements  counting  the  two  suspenders  cd,  which 
must  always  be  included  when  the  elements  alone  are  regarded.  For  Fig.  3o,  e=2  and 
for  Fig.  3p,  e=8. 

It  should  also  be  noted  that  Eq.  (3c)  is  applicable  to  any  structure  as  a  whole, 
whether  composed  of  one  or  many  elements.  This  equation  will  apply  when  the  individual 
members  and  pin  points  are  counted  and  also  when  the  elements  and  their  connecting 
points  only  are  considered. 

For  example,  in  Fig.  SN,  there  are  56  members,  32  pin  points,  and  8  reaction  con- 
ditions, making  n  =  Sr+ra  —  2p=8+56  —  2X32=0,  whence  the  structure  is  statically 
determinate.  Now  when  the  elements  alone  are  considered,  then  e=m--=Q,  p=7  and 
2r=8,  and  Eq.  (3c)  again  gives  ri  =  5>+ra-2/>=8+6-2X7=0.  Eq.  (3E)  gives 
n'  =  2r  — e—2=8—6—  2=0.  Hence,  both  equations  may  serve  to  test  the  external  redun- 
dancy, when  the  elements  alone  are  considered,  but  internal  redundancy  requires  count- 
ing all  members  and  pin  points  in  the  entire  structure.  It  is  thus  important  to  understand 
the  exact  interpretation  of  these  two  test  equations. 

As  a  general  rule  it  is  best  to  count  intersecting  web  members  as  four  members 
instead  of  two,  though  the  result  is  usually  identical.  Thus  in  Fig.  SK,  n  =  2r+m  —  2p  = 
34-17—20=3+21—24=0.  But  in  this  example  the  structure  is  not  composed  of 
a  succession  of  triangles  and  when  pairs  of  members,  meeting  in  a  point,  are  removed, 
the  structure  will  not  reduce  to  a  single  triangle,  showing  that  it  is  not  stable. 

Fig.  3cj  is  another  example  of  the  toggle  joint  when  the  members  at  c  are  not 
connected. 

The  following  table  was  arranged  to  illustrate  all  these  points  with  reference  to  the 
several  cases  represented  by  Figs.  3D  to  3n. 


Fig. 

JY 

m 

p 

e 

n  =  ~  r  +  -m  —  2p 

n1  =  £r-e-2 

Remarks. 

3D 

3 

1 

2 

1 

3+   1-  4  =  0 

3-1-2=0 

Determinate. 

3E 

6 

1 

2 

1 

6+    1-  4  =  3 

6-1-2=3 

3  times  ext.  ind. 

3r 

6 

1 

2 

1 

6+   1-  4  =  3 

6-1-2=3 

3  times  ext.  ind. 

3G 

3 

8 

5 

1 

3+   8-10=1 

3-1-2=0 

Once  int.  ind. 

3H 

4 

21 

12 

1 

4+21-24  =  1 

4-1-2=1 

Once  ext.  ind. 

3j 

6 

17 

10 

1 

6+17-20  =  3 

6-1-2=3 

Thrice  ext.  ind. 

3K 

3 

21 

12 

1 

3+21-24  =  0 

3-1-2=0 

Not  stable. 

3L 

5 

45 

24 

1 

5+45-48  =  2 

5-1-2=2 

Twice  ext.  ind. 

3M 

3 

33 

18 

1 

3+33-36  =  0 

3-1-2=0 

Determinate 

3N 

8 

56 

32 

6 

8+56-64  =  0 

8-6-2=0 

Determinate. 

3o 

4 

10 

7 

2 

4+10-14  =  0 

4-2-2=0 

Determinate. 

3p 

10 

16 

13 

8 

10+16-26  =  0 

10-8-2  =  0 

Determinate. 

3<> 

3 

9 

6 

1 

3+   9-12  =  0 

3-1-2=0 

Case  of  infinite  stress. 

3R 

8 

3 

4 

3 

8+   3-   8  =  3 

8-3-2=3 

Thrice  ext.  ind. 

CHAPTER  II 

THEOREMS,   LAWS,   AND   FORMULAE  FOR  FRAMED   STRUCTURES 

ART.  4.     ELASTIC   DEFORMATIONS,  FUNDAMENTAL  EQUATIONS 

Elastic  Deformations 

Let  S  =  total  stress  in  any  member  of  a  frame  resulting  from  any  cause,  or  causes, 

designating  tension  by  + . 
I  =  length  of  any  member  when  its  S  =  0. 
Al=  change  in  I  due  to  stress  S,  +  for  elongation. 
F  =  cross-section  of  any  member,  prismatic  in  form. 
t  =  a  uniform  change  in  temperature  in  degrees,  +  for  rise. 
£=  coefficient  of  expansion  per  degree  of  temperature. 
E  =  modulus  of  elasticity. 
f—S/F—umt  stress  in  any  member. 
JZ/Z=relative  elongation. 
l/FE  =|0=the  extensibility,  frequently  employed  for  brevity. 

Then  according  to  Hooke's  law  and  within  the  elastic  limit  of  the  material, 


EF 


E 


from  which 


SI 


(4A) 


This  equation  represents  the  elastic  deformation 
for  any  member  of  any  frame  and  is  a  fundamental 
elasticity  condition. 

Referring  now  to  Fig.  4A,  let  AB  be  any  member 
of  any  frame  which,  as  a  result  of  elastic  distortions  of 
the  frame,  is  made  to  undergo  displacements  Ji  at  A 
and  J2  at  B  and  a  change  in  its  original  length  of  Al. 
The  new  length  of  the  member  is  then  I  +  M  and  its  new 
position  may  be  shown  as  A\Bi. 

Since  the  displacements  Ji  and  A%  are  very  small 
compared  with  the  length  AB,  the  two  lines  AB  and 
A\BI  are  assumed  parallel  for  the  present  purpose. 
The  member  and  its  displacements  are  referred  to  rectangular  axes  in  Fig.  4A.  Then 


FIG.  4A. 


11 


12  KINETIC  THEORY  OF   ENGINEERING  STRUCTURES  CHAP.  II 

which,  differentiated,  gives 

2ldl=2(xb  —xn)  (dxb  -dxa)  +2(yb-ya)  (dyb  -dya)  . 

Also  from  the  figure  xb  —  xa=l  cos  a.  and  yb—ya^l  cos  /9,  which  values  substituted 
in  the  previous  equation,  dividing  through  by  21,  and  replacing  the  differentials  by 
small  finite  increments  A,  gives 


Al  =  (Axb-  Jxa)  cosa  +  (4yb-  Aya}  cosp  =  --  +  etl,    .....     (4s) 

tir 

which  is  a  fundamental  elasticity  condition  for  any  frame. 

Hence,  every  frame  will  involve  twice  as  many  of  the  unknown,  subscript-bear- 
ing displacements  Ax  and  Ay,  as  there  are  pin  points,  while  there  will  be  as  many  equa- 
tions of  the  form  Eq.  (4e),  as  there  are  members  in  the  frame. 

It  will  now  be  shown  that  any  frame,  whether  involving  redundancy  or  not,  is  capable 
of  anatysis  by  proving  the  following:  For  any  frame  in  stable  equilibrium,  there  are  as 
many  possible  condition  equations  as  there  are  unknown  quantities,  provided  the  elasticity 
conditions  Eq.  (4s)  are  included.  This  will  be  true  of  a  statically  determinate  frame 
without  including  the  elasticity  conditions. 

The  ultimate  analysis  of  any  frame  includes  the  determination  of  the  reaction  forces 
or  conditions;  the  stresses  in  all  the  members;  and  the  deformation  of  the  frame  as  a 
result  of  these  stresses.  The  deformation  is  considered  solved  when  the  displacements 
Ax  and  Ay,  of  all  the  pin  points,  are  found. 

It  is  assumed  that  for  any  structure  under  consideration,  the  externally  applied  loads 
P,  the  changes  in  temperature  t  and  the  abutment  displacements  Ar  (if  any  exist)  are 
all  known  and  that  wherever  movable  connections  or  roller  bearings  occur,  these  are 
frictioriless. 

Let  S\,  82,  83,  etc.,  be  the  stresses  in  the  several  members  meeting  in  some  particular 

pin  point  m. 
ct\,  (Xz,  «3,  etc.,  be  the  angles  which  these  members  make  with  the  x  axis  of 

coordiantes. 

/?i,  /?2,  /?3,  etc.,  be  such  angles  with  the  y  axis. 
Px=the  sum  of  the  components  parallel  to  the  x  axis,  of  all  the  external  forces 

P  acting  on  the  point  m. 

Pv=the  sum  of  the  components  parallel  to  the  y  axis,  of  all  these  forces  P. 
p,  m  and  2r  as  previously  defined  in  Art.  3.     Then 


(4c) 
cos  ,5=0 

because  for  a  state  of  equilibrium,  thje  sum  of  the  horizontal  and  vertical  components  of 
all  the  forces  acting  on  any  one  pin  point  must  respectively  equal  zero. 

Hence,  for  every  pin  point  of  a  frame,  two  equations  of  the  form  of  Eqs.  (4c)  may 
be  written,  expressing  equilibrium  of  the  internal  and  external  forces  for  that  pr'- 


ART.  4    THEOREMS,  LAWS,  AND  FORMULAE  FOR  FRAMED  STRUCTURES       13 

Also,  one  equation  of  the  form  of  Eq.  (4s)  may  be  written  for  each  member  of  a  frame. 
Besides  these,  there  will  be  one  equilibrium  equation  for  each  reaction  condition. 

Therefore,  every  stable  frame  affords  2r+m+2p  condition  equations  of  the  first 
degree,  involving  as  many  unknowns  as  there  are  equations,  thus: 

2p  equations  of  the  form  (4c)  involving    ra  unknowns  S. 
m  equations  of  the  form  (4B)  involving  2p  unknowns  Ax  and  Ay. 
Zr  equations  for  the  reactions  involving  Zr  unknowns  R. 


Total  2p-fra  +  Zr  equations,  involving  2p+m  +  £r  unknowns. 

It  follows  then  that  the  stresses  S,  the  reactions  R  of  known  direction,  and  the  Ax 
and  Ay  projections  of  the  pin-point  displacements,  may  all  be  represented  as  linear  func- 
tions of  the  horizontal  and  vertical  components  of  all  the  external  forces  P,  plus  similar 
functions  of  assumed  temperature  changes  t,  plus  linear  functions  of  the  abutment 
displacements  Ar,  Jr2,  Ar3,  etc. 

Thus  any  unknown  function  of  any  frame  may  be  expressed  by  an  equation  of  the 

form 

Z=f(Pl,P2,  P3,  etc.)+/i(0+/2(M,M,  ^3,  etc.),  .....     (4o) 


in  which  the  coefficients  are  independent  of  the  values  P,  t  and  Ar,  but  depend  on  the  lengths 
and  directions  of  the  members  also  on  E,  e  and  the  manner  in  which  the  frame  is  supported. 
The  law  of  the  summation  of  similar  partial  effects.  In  Eq.  (4o),  every  set  of  causes 
or  conditions  P,  t  and  Ar,  produces  a  partial  value  Zr  for  Z  and  the  ultimate  total  value 
of  Z=Z'  +Z"  +Z'",  etc.,  is  the  sum  of  all  the  partial  values  or  effects  resulting  from  the 
respective  sets  of  independent  causes  or  conditions.  Thus  each  effect,  such  as  a  stress, 
whether  due  to  loads,  temperature  or  abutment  displacements,  may  be  ascertained  or  invest- 
igated by  itself  and  the  sum  total  effect  Z  will  then  be  the  sum  of  the  several  similar  partial 
effects.  This  is  the  law  of  the  summation  of  similar  partial  effects  resulting  from  various 
causes  and  is  fundamental  to  the  analysis  of  all  structures  involving  redundancy. 

The  law  of  proportionality  between  cause  and  effect.  Eq.  (4o)  ,  being  true  for  any  set 
of  effects,  would  remain  true  for  any  multiple  of  these  effects.  Hence,  if  the  loads  are 
doubled  the  resulting  stresses  will  likewise  be  doubled,  etc.  Therefore,  the  law  of  pro- 
portionality holds  true  between  the  causes  and  their  effects.  Thus,  if  a  set  of  loads  P 
produces  stresses  S  in  the  members  of  a  frame  then  another_set  of  parallel  loads  P, 
acting  at  the  same  points,  would  produce  stresses  S  which  are  P/P  times  as  great  as^the 
stresses  S.  Or,  if  a  single  load  unity,  acting  on  a  point  m  of  any  frame  produces  reactions 
Ri,  and  stresses  Si,  then  a  load  Pm  acting  at  the  same  point  will  produce  reactions 
R  =  PmRi,  and  stresses  S  =PmSi. 


14  KINETIC    THEORY  OF  ENGINEERING  STRUCTURES          CHAP.  II 

ART.  5.     GENERAL    WORK   EQUATIONS   FOR   ANY   FRAME 

The  general  work  equation,  Clapeyron's  law.  A  loaded  frame  is  a  machine  in  equilibrium 
and  within  the  limits  of  proportional  elasticity  its  deformation  varies  directly  with  the 
magnitude  of  the  superimposed  loads. 

The  question  of  deformation  does  not  enter  into  statics  and  hence  a  general  and 
comprehensive  treatment  of  the  elastic  frame  must  necessarily  involve  the  principles 
of  mechanics.  , 

For  example,  given  a  frame  with  definite  loading  and  supports,  constituting  a  sys- 
tem in  external  equilibrium.  The  frame  in  turn  will  undergo  certain  deformations  which 
will  steadily  increase  in  proportion  to  the  internal  stresses  created  in  the  members  by 
the  external  forces  as  they  are  gradually  applied.  The  final  deformation  will  occur 
in  the  instant  when  the  external  forces  are  exactly  balanced  by  the  internal  stresses,  pass- 
ing into  the  condition  known  as  static  equilibrium. 

While  this  deformation  is  taking  place  the  applied  loads  travel  through  certain 
distances  which  are  the  paths  or  displacements  of  the  points  of  application  of  the  loads. 
Thus  a  certain  quantity  of  positive  work  of  deformation  is  performed  by  the  external  forces. 

The  internal  stresses  in  the  members  must  accommodate  themselves  to  the  deformed 
condition  of  the  frame  and  in  resisting  this  action  must  produce  negative  work  of  deformation. 

In  the  instant  when  static  equilibrium  is  established  between  the  loads  and  stresses 
the  positive  and  negative  work  of  deformation  produced  in  the  same  interval  of  time 
must  exactly  balance. 

Let  Ae=the  positive  or  externally  applied  work  of  deformation. 
.4;=  the  negative  or  internally  overcome  work  of  deformation. 
P=any  externally  applied  loads  including  the  reactions. 
$=the  stress  in.  any  member  due  to  the  loads  P. 
M  —  ihe  change  in  length  of  any  member  due  to  the  stress  S. 
£=the  displacement  of  the  point  of  application  of  any  force  P  measured  in 
the  direction  of  this  force. 

A  positive  amount  of  work  is  always  produced  when  the  force  and  its  displacement  act 
in  the  same  direction. 

The  product  %Pd  represents  the  actual  work  produced  by  a  force  gradually  applied 
and  increasing  from  its  initial  zero  value  to  a  certain  end  value  P,  thus  exerting  only 
its  average  intensity  during  the  entire  time  of  traversing  the  path  §  to  perform  this  work. 
Hence,  for  all  the  external  forces  acting  on  any  frame,  the  total  positive  external  work 
of  deformation  would  be  Ae=^Pd. 

Similarly  %SAl  represents  the  actual  work  in  any  member  subjected  to  a  gradually 
increasing  stress  of  end  value  S,  with  an  average  intensity  S/2  during  the  entire  time 
while  producing  a  change  in  its  length  of  JZ.  Hence  for  all  internal  stresses  in  any  frame 
the  total  negative  work  of  deformation  would  be  At-=^2$JZ. 

By  the  "  doctrine  of  the  conservation  of  energy,"  the  applied  work  must  equal  the 
work  overcome,  hence 


(5A) 
or 


ART.  5    THEOREMS,  LAWS,  AND  FORMULAE  FOR  FRAMED  STRUCTURES       15 

which  is  Clapeyron's  law  (1833),  and  may  be  stated  as  follows:  For  any  frame  of  constant 
temperature,  and  acted  on  by  loads  which  are  gradually  applied,  the  actual  work  produced 
during  deformation  is  independent  of  the  manner  in  which  these  loads  are  created  and  is 
always  half  as  great  as  the  work  otherwise  produced  by  forces  retaining  their  full  end  values 
during  the  entire  act  of  deformation. 

By  regarding  any  elastic  body  as  composed  of  an  infinite  multiplicity  of  members, 
it  is  readily  seen  that  Clapeyron's  law  applies  equally  to  frames  and  solid  web  elastic 
structures  when  properly  supported.  Clapeyron's  law  finds  extensive  application  in 
problems  dealing  with  deformations  of  framed  structures. 

Cases  involving  dynamic  impact  would  imply  a  certain  amount  of  kinetic  energy 
in  excess  of  the  applied  work  of  deformation.  That  is,  the  applied  forces  have  some 
initial  value,  greater  than  zero,  while  the  stresses  are  still  zero.  This  would  not  be  repre- 
sented by  the  work  Eq.  (OA),  wherein  Ae  —  Ai=Q,  but  would  give  rise  to  the  following 
equation  : 

Ae-Ai-jPf,     .....       .       .....        (OB) 

indicating  a  state  of  accelerated  motion  instead  of  one  of  equilibrium.  For  problems 
under  this  heading  see  Art.  16. 

The  law  of  virtual  work  will  now  be  considered.  It  is  a  general  law,  permitting  of 
more  varied  application  than  does  Clapeyron's  law. 

The  law  of  virtual  work  was  first  enunciated  by  Galilei  and  Stevin,  and  in  its  more 
general  form  by  Joh.  Bernoulli.  Lagrange  (1788)  reduced  the  law  to  an  algebraic  expres- 
sion for  which  the  following  derivation  may  be  given. 

According  to  the  general  law  expressed  by  Eqs.  (4c).  all  forces  (and  stresses)  acting 
on  a  pin  point  of  any  structure  in  equilibrium  will  have  components  parallel  to  any 
axis  and  the  sum  of  such  a  set  of  parallel  components  must  be  zero.  Calling  o:  the  angle 
which  any  force  or  stress  P  makes  with  the  axis  chosen,  then  for  any  pin  point 

2P  cosa=0. 

Now  if  a  displacement  J,  parallel  to  the  axis,  be  arbitrarily  assigned  to  this  pin  point 
and  assuming  equilibrium  to  continue,  then  the  product  of  J  with  the  sum  of  the  com- 
ponents must  still  be  zero.  This  is  equivalent  to  multiplying  the  above  equation  by 
J  to  obtain 

2(Pcosa)J=0. 

But  the  displacements  A  cos  a=d  are  the  projections  of  the  displacements  J  on  the 
directions  of  the  several  forces,  hence 

0,    .............     (5c) 


wherein  P  signifies  that  the  forces  are  in  every  sense  independent  of  the  displacements  d. 
This  is  the  law  of  virtual  work,  and  is  applicable  to  any  point  or  group  of  points  and 
hence  to  any  frame  or  group  of  bodies.  The  displacements  d  may  be  any  possible  dis- 
placements whether  or  not  subject  to  the  law  of  elastic  deformation,  provided  equilibrium 
exists. 


KINETIC    THEORY  OF   ENGINEERING  STRUCTURES 


CHAP.  II 


Stated  in  words  this  law  implies  that  for  any  point,  frame  or  body  acted  on  by  loads 
P  in  an  established  state  of  equilibrium,  the  sum  total  work  performed  by  these  forces,  in 
moving  over  any  small  arbitrary  but  possible  displacements  o,  must  always  be  equal  to  zero. 

If  the  time  element  is  introduced  into  Eq.  (oc)  giving  SPo/f  =0,  the  law  of  virtual 
velocities  is  obtained. 

Professor  Otto  Mohr's  work  equations.  Proceeding  from  Eq.  (5c),  Professor  Mohr,  in 
1874,  developed  two  equations  which  furnish  the  most  comprehensive  laws  for  framed  struc- 
tures. These  equations  are  now  derived. 

Given  any  frame  carrying  the  arbitrarily  assumed  loads  PI,  P2,  P3,  etc.,  which 
in  combination  with  temperature  changes  produce  stresses  Si,  *S2,  83,  etc.,  in  the  several 
members  of  the  frame  and  thus  constituting  a  system  of  forces  in  static  equilibrium. 

All  the  pin  points  of  the  frame  are  now  supposed  to  be  subjected  to  small,  arbitrarily 
assigned  displacements  A\,  J2,  J3,  etc.,  which  may  have  been  produced  by  some  other 
system  of  actual  loads  PI,  P2,  P3,  etc.,  entirely  independent  of  the  loads  P.  The  J's 
will  naturally  vary  in  amount  and  direction  for  each  pin  point. 

According  to  the  la\vs  for  static  equilibrium,  the  sum  of  the  components,  taken  in 
any  fixed  direction,  of  all  forces  acting  on  any  pin  point  of  a  frame  must  equal  zero. 

Choosing  for  the  fixed  direction  the  displacement  Jlr 
then  for  any  particular  pin  point,  Fig.  SA,  the  follow- 
ing equation  may  be  written: 


cos 


which  multiplied  through  by 
JiPx  cos  61  +  Ji 


5  cos  0=0; 
i  gives 
cos  0=0. 


But  from  the  figure,  A\  cos  #1  =<?i,  which  value   sub- 
stituted in  the  above  equation  gives 


0=0. 


(on) 


FIG.  SA. 


Eq.  (on)  represents  the  virtual  work  of   all  forces 
acting  on  the  pin  point  1,  in  accordance  with  Eq.  (5c). 
One  such  Eq.  (5n)  may  be  written  for   each  pin 
point  and  the  sum  of  all  these  equations  would  furnish 
the  total  virtual  work  for  the  whole  frame. 

In  this  sum  equation,  each  load  P  will  occur  only  once,  while  each  stress  S  will  occu» 
twice,  being  involved  once  for  each  end  of  the  same  member. 
The  terms  containing  the  forces  P  will  give  the  sum 


The  sum  terms  involving  the  same  stress  S  may  be  found  for  each  member  as 


cos 


AHT.  5    THEOREMS,  LAWS,  AND  FORMULA  FOR  FRAMED  STRUCTURES       17 

since  the  two  displacements  J  cos  </>  for  the  two  ends  of  the  same  member  give  the  sum 
Jl,  or  actual  change  in  the  length  of  such  member. 

Hence,  the  terms  containing  the  stresses  S  will  give  the  sum 


where  the  minus  sign  is  due  to  the  fact  that  the  stresses  applied  to  pin  points  as  external 
forces  are  always  opposite  in  direction  with  their  respective  JZ's. 
The  final  sum  equation  for  all  pin  points  then  becomes 


or 

2P3  =  2SM,    ............     (5s) 

which  may  be  said  to  represent  a  condition  of  elastic  equilibrium  as  distinguished  from 
static  equilibrium.  The  law  thus  expresses  the  equality  between  the  external  and  internal 
virtual  work  of  deformation  for  real  displacements  and  arbitrary  cases  of  loading,  provided 
elastic  equilibrium  exists. 

The  external  forces  P  necessarily  include  the  reactions  due  to  the  applied  loads  P. 
However,  the  work  of  the  reactions  is  always  zero  when  the  abutments  are  immovable. 
In  the  general  case  involving  abutment  displacements  Jr  in  the  directions  of  the  reac- 
tions R  produced  by  loads  P,  Eq.  (OE)  may  be  written 

....     (OF) 


It  should  be  repeated  that  the  displacements  d,  Ar  and  Al  are  actual  and  mutually 
dependent  on  the  same  causes,  such  as  the  actual  loads  P  and  temperature  changes. 
The  arbitrary  loads  P,  reactions  R  and  stresses  S  form  a  system  in  elastic  equilibrium 
which  is  independent  of  the  actual  loads  P,  reactions  R  and  stresses  S,  and  hence  inde- 
pendent of  the  actual  displacements. 

In  the  special  case  where  the  forces  P,  S  and  R  become  identical  with  the  forces 
P,  S  and  R,  Eq.  (5r)  becomes 

v          _  v  -         yS2* 
EF' 

which  is  in  accordance  with  Clapeyron's  law. 

Eq.  (5r)  is  the  fundamental  law  of  framed  structures  and  includes  all  conditions 
of  equilibrium  of  the  external  forces;  of  the  external  forces  and  internal  stresses;  and 
of  all  relations  existing  between  the  stresses  and  the  distortions  of  any  frame. 

Professor  Mohr's  second  work  equation  serves  the  purpose  of  determining  any  displace- 
ment dm  produced  by  any  case  of  loading.  It  follows  from  Eq.  (5r)  by  allowing  all  the 
arbitrary  loads  to  vanish  and  substituting  therefor  a  single  load  unity  and  the  stresses 
Si  and  reactions  RI  resulting  from  such  unit  load.  The  new  work  equation  then  becomes 


is 


KINETIC  THEORY   OF   ENGINEERING  STRUCTURES 


CHAP.  II 


wherein  the  unit  loading,  henceforth  called  the  conventional  loading,  may  be  a  single 
force  unity  or  a  moment  equal  to  unity.  In  the  latter  case  the  o  will  represent  a  rotation 
measured  in  arc.  By  dividing  Eq.  (5n)  by  unity  the  stresses  Si  become  abstract  num- 
bers and  the  equation  will  read 


The  applications  of  this  law,  universally  known  as  Mohr's  icork  equation,  will  be  shown 
in  the  following  articles  6  and  7.  The  work  represented  by  the  conventional  loading  will 
generally  be  designated  by  W. 

ART.  6.     DISPLACEMENTS   OF   POINTS.     STATICALLY  DETERMINATE   STRUCTURES 
BY  PROFESSOR  MOHR'S  WORK  EQUATION 

The  relative  displacement  8m  between  any  two  pin  points  mi  and  m,  of  any  frame  Fig. 
GA,  when  Al  is  given  for  each  member  of  the  frame,  may  be  found  by  applying  Eq.  (5i) . 

If  the  abutments  undergo  known  displacements  Ar,  as  a  result  of  the  given  actual 
loading,  this  effect  on  the  required  dm  must  be  included. 

Likewise  temperature  displacements  may  be  considered,  but  the  problem  will  first 
be  treated  by  neglecting  this  effect. 

Now  assume  two  unit  forces,  applied  at  the  points  mi  and  m  respectively,  and,  acting 
in  opposite  directions  along  the  line  tn\<m.  The  directions  of  these  unit  forces  should  be  so 

chosen  as  to  make  the  conventional  work  l-om  a 
positive  quantity.  This  means  that  if  dm  is  an 
elongation,  then  the  unit  forces  must  be  so 
applied  as  to  elongate  the  distance  m\m.  This 
rule  will  be  universally  applied  to  determinate 
structures  as  well  as  to  indeterminate  for 
redundant  members  or  conditions.  A  negative 
result  will  indicate  an  erroneous  assumption 
in  the  direction  of  the  unit  conventional 
load. 

The  stresses  Si  and  abutment  reactions  R i 
resulting  from  the  two  unit  loads  acting  at  mi 
and  m  are  now  determined  from  a  Maxwell  dia- 
gram or  by  computation. 

The  work  Eq.  (5n)  applied  to  this  assumed  or  conventional  loading  then  gives,  after 
substitution  of  values  Al=Sl/EF,  excluding  temperature  effect, 


FIG.  GA. 


(GA) 


The  stresses  S  are  those  produced  in  the  structure  by  any  real  case  of  loading,  as  loads 
P,  temperature  changes  or  abutment  displacements,  giving  rise  to  the  actual  changes 
Al  in  the  lengths  of  the  members. 


ART.  7  THEOREMS,  LAWS,  AND  FORMULAE  FOR  FRAMED  STRUCTURES   19 

Eq.  (6A)  then  offers  a  solution  to  the  above  problem,  because  all  the  quantities  except 
o,n  are  known  or  can  be  found  from  the  given  data. 

If  temperature  effects  are  to  be  included  then  Eq.  (4A)  gives  the  value  for  JZ,  thus 
furnishing  the  final  equation 


~+eU)-ZRlJr  .....    \     .     .     (6B) 
&V  / 

In  a  similar  manner  the  relative  displacement  of  any  pair  of  points,  the  deflection 
of  any  point,  or  the  angular  change  between  any  pair  of  lines  of  any  frame,  resulting  from 
given  changes  M  of  all  the  members  of  the  frame,  may  be  found  by  assigning  such  con- 
ventional loads  as  will  produce  the  conventional  work  1  •  d  on  the  frame  at  the  point  or 
points  in  question.  See  also  Art.  9  for  other  examples  of  this  class. 

ART.  7.     INDETERMINATE  STRUCTURES  BY  PROFESSOR  MOHR'S  WORK  EQUATION 

A  truss,  according  to  Chapter  I,  may  include  more  than  the  statically  necessary 
number  of  members  or  reaction  conditions,  and  is  then  called  statically  indeterminate. 
It  is  now  proposed  to  show  the  manner  in  which  Mohr's  work  equation  may  be  employed 
to  find  the  stresses  and  reactions  in  such  a  structure,  loaded  by  any  system  of  loads 
P  concentrated  at  the  several  pin  points  of  the  frame. 

The  structure  must  be  so  constituted  that  when  all  redundant  members  and  reaction 
forces  are  removed,  the  remaining  frame  will  represent  a  statically  determinate  structure, 
including  the  necessary  conditions  for  proper  support.  This  determinate  frame  will 
always  be  called  the  principal  frame  or  system.. 

Now  let  Xa,  Xi,  Xc,  etc.,  represent  the  stresses  produced  by  the  applied  loads  in 
any  redundant  members  or  supports,  as  the  case  may  demand.  When  these  stresses 
are  applied  to  the  principal  system,  together  with  the  external  loads,  the  resulting  stresses 
in  all  the  principal  members  will  be  identical  with  those  produced  in  these  same  mem- 
bers by  the  original  loading  of  the  whole  indeterminate  frame.  This  will  also  be  true 
of  the  deformations. 

It  follows  then  that  the  loads  P  applied  to  the  indeterminate  frame,  and  the  loads 
P  and  X  applied  to  the  principal  frame  produce  identical  stresses  and  deformations  in  the 
principal  members.  Also,  the  deformations  are  in  each  case  definitely  fixed  by  the  elastic 
changes  Al  of  the  necessary  members  of  the  principal  system.  Hence,  for  applied  loads 
only  (without  temperature  effects)  the  work  of  the  stresses  in  the  principal  frame  is 
always  greater  when  a  redundant  member  is  omitted.  The  work  which  a  redundant 
member  can  do  must  necessarily  lessen  the  work  otherwise  required  of  the  principal 
system. 

If,  then,  the  elastic  displacements  of  the  several  pin  points  representing  the  points 
of  application  of  the  redundant  forces  X,  are  found  for  the  principal  system,  these  dis- 
placements will  suffice  to  solve  one  elasticity  equation  for  each  unknown  X.  This  may 
be  done  in  two  ways:  (1)  By  applying  Mohr's  work  equation  to  the  indeterminate  sys- 
tem, and  (2)  by  finding  the  requisite  elastic  displacements  of  the  principal  system  from 
Maxwell's  law.  The  solution  by  Mohr's  work  equation  will  now  be  given. 


20  KINETIC  THEORY   OF   ENGINEERING  STRUCTURES  CHAP.  II 

The  following  definitions  of  terms  will  be  strictly  adhered  to  in  all  succeeding 
discussions. 

Let     S  =the  stress  in  any  member  of  the  principal  system. 

>S>0=the  stress  in  this  member  due  to  the  loads  P  when  the  several  redundants 

X  and  t  are  all  zero,  to  be  known  as  condition  X=Q. 
£a=the  stress  in  this  member  of  the  principal  system  when  no  load  other  than 

Xa  =  1  is  active.   Condition  Xa  =  1  . 

Sb=ihe  same  when  no  load  other  than  Xi  =  l  is  active  on  the  principal  system. 
/Sc=the  same  when  no  load  other  than  Xc  =  l  is  active  on  the  principal  system. 
£j=the  stress  in  the  same  member  caused  by  a  uniform  change  in  temperature. 
Rt=&  reaction  produced  by  a  uniform  change  in  temperature. 
Mt  =  a  moment  produced  by  a  uniform  change  in  temperature. 
Xa,  Xb,  XCJ  etc.,  the  stresses  in  the  redundant  members  or  reactions. 
R0,  Ra,  Rb  Rc,  etc.,  are  denned  like  the  S's  with  like  subscripts,  but  represent 

reaction  forces  instead  of  stresses. 
da,  <>b,  $c,  are  changes  in  the  lengths  of  the  redundant  members  Xa,  Xb,  Xc, 

respectively. 

The  forces  Xa,  Xb,  Xc,  etc.,  may  be  the  stresses  in  redundant  members  or  they  may 
be  redundant  reactions  as  in  the  case  of  fixed  arches  or  continuous  girders,  etc. 

Each  of  the  above  cases  of  conventional  loading  will  be  known  as  conditions.  Thus 
condition  X  =0  will  mean  that  all  the  redundant  conditions  are  removed,  while  condi- 
tion Xa  =  l  will  signify  that  this  force  alone  is  applied  to  the  principal  system  and  all 
other  X's,  St,  and  P's  are  removed. 

The  stress  S,  in  any  member  of  a  frame  involving  redundant  conditions,  is  a  linear 
function  of  the  loads  P,  Xa,  Xb,  Xc,  etc.,  all  treated  as  external  forces  applied  to  the 
principal  system.  This  follows  because  all  conditions  of  equilibrium  are  represented 
by  linear  equations. 

Hence,  according  to  the  law  of  the  summation  of  effects  expressed  by  Eq.  (4o)  ,  the 
general  equation  for  stress  in  any  member  of  a  truss  involving  redundancy,  would  have 
the  form 


wherein  £ax=the  stress  in  any  member  due  to  the  external  force  Xa  acting  alone  on  the 
principal  system.  Similarly  56j.=the  stress  in  the  same  member  due  to  the  external 
force  Xb  acting  alone  on  the  principal  system,  etc. 

But  by  the  law  of  proportionality  Sax=SaXa,  Sbx=SbXb  and  Scx=ScXCi  etc.     Hence, 
the  general  equation  for  any  case  of  redundancy  may  be  written  : 
for  the  stress  in  any  principal  member 

/S=*So  —  SaXa  —  SbXb  —  SCXC,      etc.,      +/S< 

+ 

for  any  reaction  of  the  principal  system 

R=R0  —  Ra  Xa  —  Rb  Xb  —  Rc  Xc,   etc.,   -\-Rt      ......     (7  A) 

and  for  any  moment  on  the  principal  system 

M=M0-MaXa-MhXb-McXc,  etc.,+M« 


ART.  7     THEOREMS,  LAWS,  AND  FORMULAE  FOR  FRAMED  STRUCTURES      21 

In  these  equations  the  quantities  S0,  R0  and  M0  are  all  linear  functions  of  the 
•externally  applied  loads  P,  while  all  the  stresses,  reactions,  and  moments  bearing 
subscripts  a,  b,  c,  etc.,  are  constants  due  to  conventional  loadings  and  are  absolutely 
independent  of  all  external  loads  P  and  X. 

The  work  of  any  member  or  reaction  is  obtained  from  Eqs.  (7 A)  by  multiplying  both 
sides  of  these  equations  by  the  deformation  which  each  sustains.  Thus,  neglecting 
temperature  effects,  the  work  equations  are 

SM=[S0-SaXa-SbXb-ScXc,  etc.]JZ 
RJr=[R0-RaXa-RbXb-RcXc,  etc.]Jr 

The  negative  signs  in  Eqs.  (7 A)  and  (7e) ,  indicate  that  the  quantities  involving  the 
redundants  X  must  always  be  of  opposite  sign  to  the  stress  S0  resulting  from  the  loads 
P,  for  reasons  above  stated  in  this  article. 

Hence  the  increment  of  work  performed  by  the  redundants  X,  as  shown  by  Eqs.  (7s), 
is  always  negative  with  respect  to  the  work  performed  by  the  stress  S0  because  the  forces 
X  are  classed  with  the  external  forces. 

The  conventional  unit  loadings  are  always  applied  in  the  opposite  direction  to  the 
forces  they  replace,  and  this  in  turn  puts  these  (negative)  unit  forces  in  the  same  direc- 
tion as  the  displacements  d  which  they  accompany.  Hence  the  work  1  •  o  is  always 
positive. 

This  is  exactly  the  same  as  the  case  illustrated  in  Art.  2,  where  the  unit  load  was 
independent  of  all  redundant  conditions. 

Therefore,  whenever  in  the  following  a  conventional  unit  load  represents  a  redundant 
condition,  the  direction  of  this  unit  load  must  be  taken  in  a  direction  opposite  to  the  force 
representing  the  stress  in  such  member.  This  is  equivalent  to  saying  that  when  a 
member  X  elongates  under  stress,  then  the  unit  load  X  =  l  must  be  so  applied  as  to  move 
the  adjacent  pin  points  apart,  and  the  converse  when  the  member  shortens  under 
stress. 

Whenever  the  direction  of  a  redundant  force  cannot  be  correctly  foreseen,  then  some 
assumption  is  made  which,  if  it  be  erroneous,  will  result  in  a  negative  value  for  such 
redundant  X. 

The  following  illustration  is  given  to  make  these  points  clear  and  to  avoid  confu- 
sion of  ideas  in  all  future  problems. 

Figures  7A  to  7E  represent  an  indeterminate  structure  involving  three  redundant 
conditions  as  may  be  verified  by  applying  Eq.  (3c)  to  the  problem  in  Fig.  7A. 

The  truss  is  supported  by  hinged  bearings  at  A  and  D  and  by  columns  at  B  and  C, 
all  of  which  may  be  subjected  to  certain  displacements  which  may  be  estimated  from 
the  conditions  of  the  problem. 

Figure  7s,  then,  represents  the  principal  system  carrying  loads  P,  when  all  redundant 
conditions  are  removed  and  shows  condition  X  =0.  The  other  figures  represent,  respec- 
tively, the  conventional  loadings  Xa  =  1 ,  Xb  =  1  and  Xc  =  1 ,  the  three  X's  being  the  redundant 
conditions.  Note  here  the  directions  which  the  unit  forces  take.  They  are  negative 


22 


KINETIC  THEORY  OF   ENGINEERING  STRUCTURES 


CHAP.  II 


with  respect  to  the  forces  they  replace,  but  in  the  same  directions  as  the  displacements 
which  their  points  of  application  undergo  as  a  result  of  the  loads  P. 

The  reaction  displacements  are  assumed  to  be  downward  by  amounts  A  A,  AB,  AC, 
and  AD  for  these  supports  respectively,  and  it  is  also  assumed  that  A  and  D  move  apart 
by  an  amount  AL.  Temperature  effects  will  be  taken  up  later  being  omitted  for  the 
present. 


GIVEN  CONDITIOM 


CONDITION  X=O. 


FIG.  7A. 


CONDITION  Xf  I. 


FIG.  7c. 


FIG.  7o. 


CONDITION  Xc-l. 


FIG.  ?E. 


The  various  reactions  in  the  following  table,  given  for  the  four  conventional  load- 
ings, are  now  found  for  the  principal  system,  which  in  this  case  is  a  three-hinged  arch. 


For  Condition. 

A 

B 

C 

D 

H* 

V       f\ 

2P(L-e) 

D          (\ 

2Pe 

AnL-P(L-2e) 

Ao           L 

.DO  —  U 

o  —  vJ 

D0        L 

H°                  2h 

,,       Uh-J:} 

•A-a  — 

V,       1 

^o  =  0 

l(L-d) 

Ba  =  v 
R,      n 

a=0 
c',      n 

L>a  —  \J 

Dh-ld 

fla- 
il 

Hh     ld- 

•Afc  —  1 

V          1 

Ab            L 
A       1'd 

•Dfe  —  U 

Bn 

<-&  —  i 

L 
D      KL-rf) 

Hb      2h 
l.d 

Ac—  1 

Ac~  L 

c  —  (J 

c  —  U 

Dc          L 

Hc"  2h 

*  Obtained  by  taking  moment  equation  about  the  middle  hinge  O,  in  a  clockwise  direction. 


ART.  7  THEOREMS,  LAWS,  AND  FORMULA  FOR  FRAMED  STRUCTURES   23 

Hence  from  Eq.  (7A)  for  reactions,  these  values  give: 

-e      L-d         d 


A=A0- 

£=0 

C=0 


and 


h  —  k  d  d 

H  =  H0  —  HaXa — HbXb — HCXC  =H0 — Xa  —  ylXb  —  FJT 

S — &0  —  &aXa  —  &bXi) — SCXC 


.     .     .     (7c) 


The  work  equation  (GA),  when  applied  to  each  of  the  three  redundant  forces  Xa,  Xb 
and  Xc  and  observing  that  the  conventional  work  1-d  in  each  case  is  positive,  gives 


=  + 1  •  AL 
-  + 1  •  AB 


(7D) 


The  values  for  ^R Ar,  in  each  of  the  Eqs.  (?D)  ,  may  also  be  written  out  from  the  above 
tabulated  quantities  for  the  three  conventional  loadings  Xa  =  1 ,  Xb  =  l  and  Xc  =  I .  They 
are: 


h 

AbAA-DbAD-HbAL 
L  —  d  .  .     d  .^      d 

-ACAA-DCAD-HCAL 

d  .        L  —  d  .         d  . 


(7B) 


It  should  be  observed  that  when  the  direction  of  a  force  is  opposite  to  the  direc- 
tion of  the  displacement  over  which  the  force  travels,  then  the  product  which  represents 
work  is  always  negative.  This  is  the  case  with  all  the  work  quantities  in  Eqs.  (7E)  .  Hence, 
when  these  are  substituted  into  Eqs.  (7o)  they  become  positive. 

The  quantities  2>SaAl,  HSbAl  and  ^SCAI  still  remain  to  be  evaluated.  Eq.  (4A)  gives 
M=Sp  +  ett,  where  S=S0—SaXa-SbXb-ScXc  from  Eq.  (7  A).  Then  by  substitution 


M=[S0-SaXa-SbXb-ScXc]p+dl, 


which  gives  the  final  value 


.     .     .     (7c) 


24  KINETIC  THEORY  OF   ENGINEERING   STRUCTURES  CHAP.  IT 

Similar  expressions  follow  for  2S&JZ  and  2$CJZ,  and  these  substituted  in  Eqs.  (7o) 
give  for  the  values  d: 


da  = 

.     .     (7n) 
dc  = 


wherein  the  summations,  except  those  of  the  reactions,  include  only  the  members  of 
the  principal  system. 

Now  da,  £5,  and  dc  being  the  changes  in  the  lengths  of  the  three  redundants,  these  may 
be  evaluated  from  Eq.  (4A)  in  terms  of  the  lengths,  areas  and  stresses  of  these  members 
(or  reactions)  and  become 


-A-  ,  ,  •  ,  . 

(73) 


Hence  all  the  terms  in  Eq.  (7n)  are  now  known  except  the  three  redundant  forces 
Xa,  Xb  and  Xc,  and  having  three  elasticity  equations  involving  only  these  unknowns, 
the  latter  may  be  found  by  solving  Eqs.  (7n)  for  simultaneous  values  of  the  X's,  with 
the  aid  of  Eqs.  (?E)  and  (7j) . 

It  is  thus  seen  that  the  abutment  displacements  are  of  vital  importance  in  determin- 
ing the  magnitudes  of  the  stresses  in  any  statically  indeterminate  structure.  When- 
ever these  displacements  cannot  be  determined  with  any  degree  of  certainty,  or  when 
small  displacements  indicate  large  resulting  stresses,  then  such  structures  should  not 
be  built.  This  applies  particularly  to  fixed  arches  and  continuous  garders. 

Eqs.  (7n),  based  on  Professor  Mohr's  work  equation,  thus  furnish  a  means  for 
the  analysis  of  any  statically  indeterminate  structure. 

ART.  8.  INDETERMINATE  STRUCTURES  BY  MAXWELL'S  LAW 

The  application  of  Maxwell's  law  to  the  same  analysis  will  now  be  given  by  express- 
ing the  summations  in  Eqs.  (7n)  in  terms  of  the  elastic  deformations  of  the  structure 
instead  of  the  stresses  in  the  members. 

Letoma=the  displacement  of  the  point  of  application  m  of  any  load  P,n,  in  the 
direction  of  this  load,  when  the  principal  svstem  is  loaded  with  only 

Xa  =  l. 

dmb=fi  similar  displacement  of  the  same  point  m  for  the  conventional  load- 
ing X6  =  l. 

dnlc=ihe  same  for  condition  Xc  =  l. 

daa  =the  change  in  length  of  the  member  a  for  condition  Xa  =  l. 
d^  =the  change  in  length  of  the  same  member  a  for  condition  Xb  =  l. 
doc  =the  similar  change  for  condition  Xc  =  \. 
dba  =  the  change  in  length  of  the  member  b  for  condition  Xa  =  1 . 
dbb  =the  change  in  length  of  the  member  6  for  condition  Xb  =  l. 


ART.  8    THEOREMS,  LAWS,  AND  FORMULA  FOR  FRAMED  STRUCTURES       25 

dbc  =  the  change  in  length  of  the  member  6  for  condition  Xc  =  l. 
dca  =the  change  in  length  of  the  member  c  for  condition  Xa  =  l. 
dct,  =the  change  in  length  of  the  member  c  for  condition  X&  =  1. 
dcc  =the  change  in  length  of  the  member  c  for  condition  Xc  —  l. 
oal  =the  change  in  length  of  the  member  a  resulting  from  a  change  in  tempera- 
ture t°,  when  the  principal  system  is  otherwise  not  loaded,  hence 

x=o,  p=o. 

Ob<=the  same  for  the  member  b. 
dct  =the  same  for  the  member  c. 

For  the  sake  of  simplicity,  the  abutments  will  be  assumed  immovable,  thus  making 
all  Jr=0.  This  part  of  the  previous  problem  remains  unchanged. 

The  work  Eq.  (OE)  for  the  condition  X— 0,  (when  the  loads  P  only  are  acting  on 
the  principal  system  and  producing  the  stresses  S0)  now  becomes  for  the  three  cases  of 
displacements 


> 


Similarly  applying  the  work  Eq.  (GA)  to  the  conditions  Xa  =  l,  Xb  =  l  and  Xe  =  l, 
the  various  displacements  become 

1  •  daa  =  2SaMu;  l-3ab  =  2SaMb  and  1  •  3ac  =  2SaMc; 
and  by  inserting  values  for  Alaj  4lb  and  Alc  from  Eq.  (4A)  then 


and  similarly  for  conditions  Xb  =  1  and  Xc  =  1 


2  ScSap  =dca]   S  ScSi,p  =dcb',  2  ScScp  =  dc 


(SB) 


Finally  the  work  Eq.  (6B),  for  temperature  displacements  for  each  of  the  conven- 
tional unit  loadings,  gives 

SS0etf  -  1  •  3at  ;  2Sbdl  =  l-3bt  and  SScstf  =  1  •  dct  ......     (8c) 

Substituting  all  these  values  from  Eqs.  (SA)  ,  (SB)  ,  and  (8c)  into  Eqs.  (?H)  ,  the  follow- 
ing important  equations  are  obtained  : 

—Xcdac  —  S#aJr  +dat 


Ob  =  ZPmdmb  —Xadba  —Xbdbb—Xcdbc  —  T>Rb 
dc  =  2Pm3mc  -Xadm  -Xhdcb-Xc3cc  -  2Rc 


(SD) 


It  should  be  noted  that   for  the  quantities  on  the  right-hand  side  of  Eqs.  (So),  the 
single  subscripts  and  the  second  one  of  the  double  subscripts  always  refer  to  the  con- 


26  KINETIC  THEORY   OF  ENGINEERING  STRUCTURES  CHAP.  II 

ventional  loadings  Xa  =  l,  Xb  =  l  and  Xc  =  l,  while  the  first  of  the  double  subscripts  always 
refers  to  a  point  or  member  of  the  structure. 

According  to  Maxwell's  law  (proven  in  Art.  9)  the  following  equalities  exist  in  Eqs.  (So)  : 

$ab  =  dba',    dac=^ca  and  dbc  =  Ocb. 

This  means  that  the  order  of  the  subscripts  may  be  interchanged  at  will  without 
altering  the  equations,  thus  greatly  simplifying  the  solution  of  problems. 

All  of  the  displacements  d,  in  Eqs.  (So),  may  now  be  determined  by  four  Williot- 
Mohr  displacement  diagrams  (see  Chapter  VI),  drawn  respectively  for  the  conditions 
Xa  =  l,  Xb  =  l,  Xc  =  l  and  t  =  l  acting  on  the  principal  system.  Hence,  the  three  redun- 
dant conditions  X  are  again  found  by  solving  Eqs.  (SD)  for  simultaneous  values. 

It  is  thus  seen  that  any  indeterminate  structure  may  be  analyzed  in  either  of  two 
ways,  by  Eqs.  (7n)  or  (SD),  according  to  the  choice  of  the  designer  or  the  nature  of  the 
problem. 

The  redundants  X  being  found  by  either  of  the  above  methods,  all  the  stresses  S 
and  reactions  R  may  now  be  determined  from  Eqs.  (7  A)  .  In  these  the  X's  are  the  redun- 
dant forces  from  Eqs.  (7n)  or  (So)  and  the  stresses  S0,  Sa,  Sb  and  Sc  are  those  found  for 
the  conventional  loadings  on  the  principal  system  and  are  independent  of  the  values 
X  and  of  each  other. 

The  summations  in  all  the  previous  equations  include  only  the  members  of  the 
principal  frame.  However,  an  equation  of  the  form  (7n)  may  be  written  to  cover  all 
members,  including  the  redundant,  and  this  form  is  frequently  very  useful. 

The  work  Eqs.  (7o)  when  made  to  cover  all  members  of  an  indeterminate  frame 
become 

S#cJr  =  SScJZ  .....     .     (SE) 


These  values  inserted  into  Eq.  (7o),  and  others  of  that  form,  give 

+  2Saea  1 

(8r) 


2#c  Ar  =  2ScS0p  -Xa^ScSaf)  -XbZScSbf)  -  Xc^S 

wherein  the  summations  extend  over  all  the  members  including  the  redundant,  and  all 
the  terms  retain  their  previous  significance.  Therefore,  when  Xa  =  l,  then  S0  =0, 
Sa  =  l=Xa,  Sb  =0,  and  Sc  =0. 

The  first  terms  of  the  right-hand  side  of  each  of  the  Eqs.  (8r),  according  to  Eqs. 
(SA),  may  be  expressed  in  terms  of  the  external  loads  as  follows: 

2iS0Sap=  2Pm$mo;    ^S0Sbf>~^Pmdmb',    2)S0*STCjO=lIPm^7nc,  .     .     .     (SG) 

and  may  be  evaluated  from  Williot-Mohr  displacement  diagrams,  Chapter  VI.  All  the 
other  summations  may  be  determined  once  for  all  either  from  Maxwell  stress  diagrams, 
or  in  terms  of  displacements  by  employing  Eqs.  (SB)  . 

Problems  of  the  kind    treated  in  Art.  6  can   now  be  solved  for  any  indeterminate 
structure  by  following  precisely  the  same  method  there  indicated  except  that  the  redundants 


ART.  o    THEOREMS,  LAWS,  AND  FORMULA  FOR  FRAMED  STRUCTURES       27 

must  first  be  found  by  one  of  the  two  methods  just   given,  finally   employing  Eq.  (?A), 
to  ascertain  the  stresses.     These  stresses  being  found,  the  elastic  changes  in  all 
members  are  computed.      Then  by  applying  a  conventional  unit  load  to  the  point  for 
which  the  displacement  is  desired,  determine  the  stresses  in  the  principal  system    for 
this  unit  load  and  proceed  as  before  by  using  Eq.  (6A). 

It  may  be  well  to  mention  in  closing  this  subject  that  there  is  usually  a  wide  latitude 
n  the  choice  of  the  redundant  conditions.  Thus,  in  Fig.  ?A,  one  abutment  might  have 
>een  placed  on  rollers,  thus  making  the  principal  system  a  simple  truss  on  two  supports^ 
Then  Xa  would  have  become  the  horizontal  thrust  instead  of  the  stress  in  a  member 

ART.  9.     PROF.   MAXWELL'S   THEOREM    (1864) 

This  theorem,  known  as  the  law  of  reciprocal  displacements,  and  previously  men- 
tioned in  discussing  Eqs.  (8D),  establishes  the  mutual  relation  between  the  elastic  dis- 
placements of  a  pair  of  points,  or  a  pair  of  lines,  whenever  these  displacements  result 
from  simultaneous  conditions  of  loading ;_  provided  that  the  arrangement  of  the  membe 
remains  unchanged  and  the  supports  are  immovable. 

For  the  sake  of  simplicity  it  is  assumed  that  the  frame  is  in  a  condition  of  no  stress 
and  that  there  are  no  temperature  changes. 

Clapeyron's  law  then  applies  and  the  equation  for  actual  work  becomes 


wherein  the  P's  are  concentrated  loads,  and  the  3's  are  the  deflections  of  the  points  on 
which  the  loads  act  in  the  respective  directions  of  these  loads. 


FIG.  QA. 


FIG.  OB. 


Each  of   the  products  \Pmdm  may  now  be  regarded    as  a  summation  of  work  pro- 
duced by  some  group  of  loads  such  that  the  work  of 
the  group  is  exactly  identical  with  the  work  rep- 
resented by   %Pmdm.      Figs.   QA,    9s,    and   9c   will 
illustrate  just  what  is  meant  by  such  groups  of  loads. 

If  in  Fig.  OA,  two  equal  and  opposite .  forces 
Pm  are  applied  at  the  points  ml  and  m,  then  the 
resulting  om  represents  the  relative  displacement 
between  these  two  points.  We  call  this  case 
the  loading  of  a  pair  of  points  corresponding 
to  the  case  illustrated  in  Fig.  GA,  where  the  loads  P  are  each  equal  to  unity. 


FIG.  9c. 


28  KINETIC  THEORY  OF  ENGINEERING  STRUCTURES          CHAP.  IT 

Fig.  OB  shows  a  pair  of  equal  and  opposite  forces  perpendicular  to  a  line  ik  and 
producing  a  couple  of  moment  1-Pm.  This  represents  the  loading  of  a  line.  The  result- 
ing dm  represents  the  angular  change  in  the  line  ik  expressed  in  arc.  This  case  may  be 
designated  as  the  loading  of  a  line  m  and,  whenP  =  l,  it  becomes  the  unit  loading  of  this 
line. 

Fig.  9c  represents  the  loading  of  a  pair  of  lines  m^  and  m.  The  dm  of  this  group 
of  loads  is  the  relative  angular  change  between  the  two  lines,  or  it  is  the  change  in  the 
angle  6,  which  results  from  the  conventional  loading. 

The  three  groups  of  loadings  are  typical,  and  any  particular  load  applied  to  a  frame 
may  become  one  load  of  such  a  group,  but  not  of  two  or  more  groups  simultaneously. 

For  brevity,  any  such  group  of  loads  will  be  known  as  a  loading  and  the  corresponding 
o  will  be  the  path  or  elastic  displacement  of  the  loading. 

The  several  d's  are  always  linear  functions  of  the  applied  loads  P  and  can  be  expressed 
as  follows: 


•   •  •  dbmPm;   \ (9B> 

O  m  ~  0  I>M r a  ~r  0 rnb* b  ~r   •    •    •    OmlnLn 


wherein  the  d's  with  the  double  subscripts  are  independent  of  the  loads  P,  and  for 
example  oam  represents  the  special  value  of  da  when  Pm  =  1  and  all  the  other  loads  P 
are  zero,  all  in  accordance  with  the  nomenclature  employed  in  Art.  8. 

Now  apply  loads  Pm  to  any  frame,  producing  stresses  Sm  in  the  several  members, 
and  changes  in  their  lengths  Jlm=Sml/EF. 

Likewise  for  a  system  of  loads  Pn,  producing  stresses  Sn,  and  changes  Aln  =  Snl/EF. 

Let  d'mn  =  the  value  of  dm,  for  the  point  m,  when  certain  loads  Pn  only  are  active. 
Also,  let  d'nm  =  the  special  value  of  dn  for  the  point  nj  when  a  certain  set  of  loads  Pm  only 
are  active.  Then  note  that  d'mn  =  dmn  when  2Pn=l  and  o'nm  =  dnm  when  2PTO=1. 

The  work  equation  for  the  loads  Pm,  stresses  Sm  and  displacements  d'mn  and  Aln  due 
the  loads  Pn  only,  is  according  to  Eq.  (OE) 

S  I 

2jPmO  mn  =  ^owz  Jtn==  Lo^-pp. 

Similarly 


S  I 

TiPnd'nm  =  2}Sndlm  ~  ^^n^W> 

therefore, 


(9c) 


which  is  Betti's  law  (1872)  and  which  extends   Maxwell's  law  to  the  summation  of  all 
members  of  a  structure. 

If  instead  of  systems  of  loads  Pm  and  Pn  as  above,  only  the  unit  loads  Pm  =  1  and 
Pn  =  1  are  successively  applied  to  the  frame,  then  Eq.  (9c)  becomes 

dmn  =  8nm,    .............     (9D) 

which  is  Maxwell's  law  (1864)  and  may  be  stated  thus: 


ART.  10  THEOREMS,  LAWS,  AND  FORMULA  FOR  FRAMED  STRUCTURES  29 

1.  The  relative  displacement  dmn  of  a  pair  of    points  TOI  and   TO,  resulting  from  a 
unit  loading  of  another  pair  of  points  n\  and  n,  is  equal  to  the  relative  displacement  onm 
of  the  pair  of  points  n\  and  n,  caused  by  a  unit  loading  of  the  pair  of  points  m\  and  TO. 

2.  The  same  equality  exists  between  the  relative  angular  changes  of   two  pairs  of 
lines,  successively  loaded  with  a  unit  loading. 

3.  The  same  equality  also  exists  between  the  linear  change  in  a  pair  of  points  result- 
ing from  a  unit  loading  of  a  pair  of  lines,  and  the  angular  change  (expressed  in  arc)  of 
the  pair  of  lines,  resulting  from  a  unit  loading  of  the  pair  of  points. 

The  practical  value  of  Maxwell's  law  when  applied  to  redundant  conditions  was  shown 
in  Art.  8.  Its  application  to  displacement  influence  lines  is  given  in  Chapter  VIII. 

ART.  10.     THEOREMS   RELATING   TO   WORK   OF   DEFORMATION 

(a)  Menabrea's  law  (1858),  or  theorem  of  least  work.  Given  a  statically  indeterminate 
framework  in  an  initial  condition  of  no  stress  for  which  the  temperature  is  known  and 
remains  constant;  also,  assuming  that  the  supports  are  either  rigidly  fixed  or  permit 
frictionless  movements,  such  that  2.RJr  =  0  for  the  entire  structure,  then  the  changes 
in  the  lengths  of  the  members  are  expressed  by  the  formula  M  =  Sl/EF,  and  the  work 
equations  for  the  several  statically  indeterminate  quantities  Xa,  Xf,,  Xc,  etc.,  are,  accord- 
ing to  Eqs.  (SB)  , 


EF 


etc., 


the  value  S  in  which  is  given  by  Eq.  (7A.)  as 

S  =  S0  —  SaXa — SbXf,  —  SCXC,  etc. 
The  partial  differentiation  of  S  with  respect  to  Xa  gives 


likewise  for  the  other  X's, 


(10B) 


(lOc) 


The  actual  work  of  deformation  for  the  entire  frame,  including  redundant  members 
as  external  forces,  becomes  by  Clapeyron's  law  or  Eq.  (5A) 


(10D) 


30  KINETIC  THEORY  OF  ENGINEERING  STRUCTURES          CHAP.  II 

which  when  differentiated  gives 

_ 

and  by  dividing  through  by  ~dXa  this  becomes,  after  substitution  of  values  from  Eqs.  (lOc) 

- 

- 


EF  EF   ' 

which  by  Eq.  (10A)  must  equal  zero.     Hence, 


,    .    ..    .        dA  .     9  A  ,       . 

=  0     and  similarlv     ^-^-=0     and     ^^=0,      ....     (10E) 


which  proves  that  the  redundant  or  indeterminate  conditions,  reduce  the  actual  work  of  defor- 
mation of  the  frame  to  a  minimum.  This  is  the  theorem  of  least  work. 

(b)  Castigliano's  law  (1879),  or  derivative  of  the  work  equation.  This  law  deals 
with  the  displacement  of  the  point  of  application  of  a  force. 

Any  external  load  Pm  acting  on  a  framework  will,  by  Clapeyron's  law,  produce  the 
actual  applied  work  of  deformation 

A=$Pmdm,     .........     .     .     (10F) 

when  dm  is  the  displacement  of  the  point  of  application  of  Pm  in  the  direction  of  Pm 
and  within  range  of  proportionality,  or  within  the  elastic  limit  of  the  material. 

Also,  the  actual  internal  work  of  deformation,  for  the  entire  frame,  becomes,  accord- 
ing to  Eq.  (10D) 


The  partial  differential  derivative  of  A  with  respect  to  any  certain  external  load 
Pmis 

'S2l\    .^sdS_  J_ 

3Pm  EF' 


wherein  3*S/3POT  is  the  derivative  of  the  stress  S  in  any  member  of  a  determinate  or 
indeterminate  structure. 

The  general  expression  for  S,  from  Eq.  (lOs),  by  substituting  for  S0  its  equivalent 
value  in  terms  of  external  loads  becomes 


,  etc.,  +SmPm-S0Xa-SbXb-ScXc,  etc.,    .     .     .     (10j) 

in  which  Si  is  the  stress  in  the  member  S  for  PI  =1,  while  Sa  is  that  stress  when  Xa  =  l, 
etc.;  finally  S  is  the  stress  in  any  particular  member  caused  by  the  combined  effects 
of  all  P's  and  Z's. 

The  several  values  Si  to  Sm  are  thus  independent  of  the  loads  P  and  X.  Also,  the 
loads  P  and  X  are  independent  of  each  other.  Hence,  S  may  be  partially  differentiated 
with  respect  to  Pm  or  X  and  Eq.  (10j)  when  so  treated  gives 

--=Sm     and     ''==~s°>  etc  ........     (10K) 


ART.  11     THEOREMS,  LAWS,  AND  FORMULA  FOR  FRAMED  STRUCTURES     31 
Substituting  this  value  in  Eq.  (lOn)  then 


but,  according  to  Mohr's  law,  Eq.  (GA),  when  abutment  displacements  are  zero,  then 

•    •  •     (10M) 


In  Eqs.  (IOL)  and  (10M)  the  summation  extends  to  all  the  members,  including  the 
redundant  conditions,  and  hence 

7)4 

Ldm=-,     .......       ........       (ION) 


which,  expressed  in  words,  means  that  the  path  om  of  a  load  Pm  is  equal  to  the  partial 
differential  derivitave  of  the  actual  work  of  the  frame  with  respect  to  Pm. 

Castigliano's  law  thus  expressed  is  equivalent  to  Mohr's  work  equation,  that  is, 
leading  to  the  identical  result  by  a  more  circuitous  process. 

In  similar  manner  the  same  law  may  be  deduced  for  girders  with  solid  webs. 

Mohr's  law  thus  offers  the  most  direct  method  of  finding  any  displacement  which 
may  result  from  any  specific  cause.  Castigliano's  law  will  lead  to  the  same  conclusions 
by  a  somewhat  less  direct  method. 

ART.  11.     TEMPERATURE  STRESSES  FOLLOW  MENABREA'S  AND  CASTIGLIANO'S 

LAWS 

When  a  structure  is  subjected  to  a  uniform  change  in  temperature  t  then,  from 
Eq.  (4A), 


This  value  of  M  when  substituted  into  Eq.  (10D)  for  Sl/EF,  gives 

;     ........     (llA) 


and  the  differential  of  A  with  respect  to  S  is 

(llB) 


Similarly  when  the  temperature  effect  is  introduced  into  Eqs.  (10A)  then  for  immovable 

abutments  as  before 

y  c  cf/ 

2S^l==+8a<&-Q  .........     (He) 


32  KINETIC  THEORY   OF  ENGINEERING  STRUCTURES  CHAP.  II 

Also,  Eqs.  (lOc)  apply  equally  when  temperature  effects  are  included,  hence  by  divid- 
ing Eq.  (lie)  by  3A"a,  and  substituting  values  from  Eqs.  (lOc),  then  Eq.  (11s)  becomes 


i 

9  A 


a~   EF 
But,  by  Eq.  (lie)  this  expression  is  equal  to  zero,  hence,  as  before 

"dA  3A 

o^r-  =  0     and  similarly     ~^-  =0  etc.,     ......     (HE) 

oXa  oXb 

which  proves  Menabrea's  law  applicable  to  the  general  case  of  stress  from  temperature 
and  applied  loads,  provided  the  abutments  are  rigid  and  immovable. 

The  introduction  of  the  temperature  factor  SefcS/  from  Eq.  (!!A)  into  Eqs.  (10o), 
to  (ION)  will  suffice  to  prove  that  Castigliano's  law  also  applies  to  the  general  case  includ- 
ing temperature  effects.  It  is  not  deemed  necessary  to  repeat  the  transformations  here. 

ART.  12.     STRESSES   DUE   TO   ABUTMENT   DISPLACEMENTS 

Abutment  displacements  produce  stresses  which  follow  Menabrea's  and  Castigliano's 
laws. 

This  is  readily  seen  when  it  is  considered  that  the  supporting  elements  may  always 
be  replaced  by  linked  members  which  take  up  these  displacements  and  undergo  elastic 
deformations.  The  immovable  supports  are  then  outside  of  these  connecting  links  as 
illustrated  in  Figs.  ID,  IE  and  IF,  and  the  links  are  counted  with  the  structural  members. 
If  then  the  abutment  displacements  are  defined  as  distortions  Jr  in  these  several 
links,  which  are  in  every  way  equal  to  the  former  displacements,  the  resulting  effect  on 
the  structure  remains  unchanged. 

Therefore,  when  the  distortions  Jr  must  be  considered  in  the  computation  of  the 
X's  and  d's,  it  is  only  necessary  to  extend  the  work  equations  to  include  these  links.  It 
is  also  clear  that  these  links  may  have  any  desired  sections,  lengths  or  values  of  E,  as  may 
be  required  for  metal  or  masonry  supports. 

Thus,  let  r  be  the  length  of  a  supporting  link; 
R  the  load  which  the  link  carries; 
Ar  the  change  in  length  of  this  link; 
Fr  the  cross-section  of  this  link; 
er  the  coefficient  of  expansion  ; 
t  the  change  in  temperature  from  the  normal; 
Er  the  modulus  of  elasticity  for  any  material. 
then  from  Eq.  (4  A) 


The  general  work  Eq.  (11  A)  then  becomes 

V027 


ART.  12    THEOREMS,  LAWS,  AND  FORMULA  FOR  FRAMED  STRUCTURES     33 

wherein  Ar  may  be  regarded  as  a  constant  and  Menabrea's  and  Castigliano's  laws  again 
apply. 

For  the  case  of  one  external  indeterminate,  R  becomes  Xa  and  Ar  becomes  oa,  then 
from  Eq.  (12n) 


By  differentiating  for  8  with  respect  to  Xa  and  imposing  the  condition  for  minimum 
this  becomes 

s      3S     3  n9  v 

:t         +  da~ 


Also,  S=S0—SaXa  and  its  differential  is 


Substituting  the  value  from  Eq.  (12o)  into  Eq.  (12c)  and  solving  for  Xa  then  the 
latter  becomes 

2S0Sa~  +  2dlSa-3a 
Xa=  ~'     ........     (12E) 


EF 

which  might  be  used  as  the  expression  for  the  horizontal  thrust  of  a  two-hinged  arch 
where  Xa  is  the  redundant  thrust. 


CHAPTER    III 


THEOREMS,   LAWS,  AND  FORMULA  FOR   ISOTROPIC  SOLIDS 

ART.  13.     GENERAL   WORK  EQUATIONS 

In  the  previous  chapters  no  consideration  was  given  to  solid  web  structures  or  other 
isotropic  solid  bodies  only  in  so  far  as  was  necessary  in  demonstrating  the  general  laws 
and  theorems  of  framed  structures. 

While  it  is  true  that  the  foregoing  discussion  is  generally  applicable  to  all  isotropic 
solid  bodies  which  are  supported  in  any  of  the  ways  given  in  Chapter  I,  and  stressed 
Avithin  the  elastic  limit  by  externally  applied  loads,  yet  the  formula?  previously  given 
will  require  some  modifications  to  better  adapt  them  to  solid  web  and  other  massive 
and  homogeneous  structures.  Also,  the  introduction  of  shearing  stress  now  enters  as 
a  further  complication. 

As  was  previously  mentioned,  all  structures  which  are  purely  isotropic  «an  involve 
only  external  redundancy.  It  is,  therefore,  desirable  to  take  up  this  subject  to  the 

extent,  at  least,  of  giving  the  special  formu- 
lae applicable  to  any  solid  in  which  the 
physical  properties  of  the  material  are  pre- 
sumably uniform  in  any  and  all  directions. 
Such  bodies  are  called  isotropic  solids. 

It  will  scarcely  be  necessary  to  prove 
at  length  all  of  the  theorems  and  laws  pre- 
viously given  for  frames,  but  a  passing  refer- 
ence at  the  proper  time  will  be  deemed  suf- 
ficient proof  of  their  general  acceptance. 

The  elastic  deformation  resulting  from 
given  stresses  in  an  infinitesimally  small 
particle  of  a  body  is  a  certain  and  definitely 
measurable  quantity  and  is  dependent  only 
on  the  magnitude  of  the  given  stresses  and 
the  physical  properties  of  the  material. 

If  then  Fig.  13A  represents  a  small  parallelepiped  of  mass,  referred  to  axes  X,  Y 
and  Z,  and  having  dimensions  dx,  dy  and  dz,  coincident  with  these  axes  respectively, 
then  the  stress  acting  at  the  corner  m  is  determined  in  magnitude  and  direction  by 
nine  components,  viz.,  three  normal  stresses  and  six  tangential  stresses  acting  in  the  three 
coordinate  planes  and  positive  in  the  directions  of  the  arrows. 


FIG.  ISA. 


ART.  13     THEOREMS,  LAWS,  AND  FORMULA  FOR  ISOTROPIC  SOLIDS  35 

Let  fx,  fy  and  fz  be  the  unit  normal  stresses  (tension  or  compression)  in  the  direc- 
tions X,  Y  and  Z,  respectively.  Also,  let  -xy  and  rxz  be  the  unit  tangential  stresses  in 
the  YZ  plane;  ryx  and  ryz  those  in  the  XZ  plane;  and  -zx  and  ~zy  those  in  the  XY  plane. 
The  first  subscript  indicating  the  normal  stress  to  which  they  belong,  and  the  second 
subscript  referring  to  the  axis. 

The  two  unit  tangential  stresses  intersecting  in  any  one  point  are  equal,  in  each 
case,  otherwise  the  body  would  rotate  about  its  center  of  gravity,  which  is  the  intersec- 
tion 0  of  the  three  normal  unit  stresses.  Also,  because  the  three  normal  stresses  and 
four  tangential  stresses,  ~,fz,  rxz,  r,y  and  TZX,  cannot  have  any  moment  about  such  a  gravity 
axis  OZ',  hence,  the  moment  of  the  two  remaining  stresses  -cyj.  and  rxy  must  equal  zero. 
But,  the  latter  having  equal  lever  arms  must  be  equal  to  each  other. 
Therefore, 


The  resulting  stress  at  the  point  m  is  then  determined  by  the  three  normal  stresses 
fx,fy,fz  and  the  three  tangential  stresses  rx,  TU,  and  TZ  in  Eqs.  (13A),  and  these  in  turn 
determine  the  deformation  of  the  parallelepiped. 

Let  Adx,  M,y  and  Adz  represent  small  elastic  changes  which  the  lengths  dx,  dy  and 
dz  undergo,  then  the  forces  represented  by  the  unit  normal  stresses  /  will  produce  the 
following  virtual  work  : 


=fxdydzJdx+fydzdxJdy+fzdxdyJdz  =    fx+ 

Also,  let  j-x,  fy  and  fz  be  the  tangents  of  small  angular  distortions  of  the 
parallelepiped  such  that  +fx  is  an  increase  in  the  angle  between  the  Y  and  Z  axes;  fv 
the  change  in  the  angle  between  the  X  and  Z  axes,  etc.  Then  the  virtual  work  of  the 
tangential  forces  is  found  thus  :  For  the  distortion  of  the  YZ  plane,  the  path  is  fxdy  and 
the  force  is  rxdxdz,  and  since  the  vertical  force  r^jjan  do  no  work  in  this  direction  then 
the  work  of  the  tangential  stress  for  the  plane  XZ  becomes  rxdxdz^rxdy,  and  similarly 
for  the  other  two  planes,  whence  the  total  work  becomes 

x  ......     (13c) 


The  expressions  Adxjdx,  4dy/dy,  and  4dz/dz  in  Eqs.  (13B)  are  rates  of  elongation, 
or  coefficients,  for  which  the  values  ax,  av  and  az  may  be  substituted.  Then  by  making 
dxdydz  =dV,  and  combining  Eqs.  (13s)  and  (13c),  the  total  internal  virtual  work  is  obtained. 
Also,  since  by  Clapeyron's  law,  this  must  be  equal  to  the  virtual  work  of  the  externally 
applied  forces,  then  the  fundamental  work  equation  for  isotropic  bodies  becomes: 

W  = 

This  equation  is  applicable  to  any  case  of  related  displacements  and  elastic  deforma- 
tions o  dr,  a,  and  7-,  so  long  as  these  are  small  in  comparison  with  the  dimensions  of  the 
structure.  The  external  forces  must  include  dead  loads  and  all  frictional  resistances 
which  may  be  active  at  points  of  support. 


36. 


KIXETIC  THEORY   OF  ENGINEERING  STRUCTURES         CHAP.  Ill 


As  was  previously  shown  for  indeterminate  frames,  so  also  the  various  elements 
in  Eq.  (13D)  may  now  be  expressed  MS  linear  functions  of  the  applied  loads  and  any 
redundant  conditions,  thus: 

R=R0—RaXa—RbXb,  etc. 

J  x  =J  xo  J  xa-A-a  J  xb-^-bj  CtC. 
fy  =fyo—fyaXa  ~fyb^b,  etc. 
fz  =^fzo  —fiaXa  —fzbXb,  etc. 
?x  =  Tro  —  ~raXa  —TxbX.b,  etc. 

etc.,  etc. 

Also,  the  virtual  work  of  the  reactions  for  each  of  the  conventional  loadings  becomes, 
(see  Eqs.  (?E)): 


(13E) 


xaf 


(13P) 


f 

2Rb  Ar  =J  [fxb( 

etc. 
Then  by  writing  Eq.  (13D)  for  a  load  Pa  =  l  the  following  is  obtained: 

-'ZRa<lr.  .     .     (13c) 


from  which  problems  of  the  kind  described  in  Art.  6  may  be  solved. 

Since  the  redundants  X  in  Eqs.    (13E)  may  be  treated  as  independent  variables, 
their  differentiation  furnishes 


_      f.-t 

a' 


"dX  a'      3Y 

which  by  substitution  into  Eqs.  (13r)  give 


etc.,  and 


~  ~r 


etc. 


'  '   (13H> 

Similar  expressions  result  for  2/S^Jr,  etc. 

The  general  work  equation  is  now  found  by  inserting  for  the  actual  distortions  a 
and  f  their  values  in  terms  of  stresses  and  the  moduli  of  elasticity  for  direct  and  tan- 
gential stress. 

The  length  dx  subjected  to  the  unit  stress  fx  and  a  rise  in  temperature  of  t  degrees 
will  be  changed  by  an  amount 


ART.  13     THEOREMS,  LAWS,  AND  FORMULA  FOR  ISOTROPIC  SOLIDS  37 

The  two  other  unit  stresses/,  and  fz  will  diminish  the  effect  on  ax  by  an  amount 

/*+/. 


mK  ' 


wherein  m  is  the  "  Poisson  number  "  (1829)  which  is  given  as  3.33  for  structural  steel 
and  3  5  for  hi-h  steel.     Professor  F.  E.  Turneaure  gives  0.1  to  0.125  for  concrete, 
quantity  m   is°  also  denned   as  the   ratio   of  lateral  to  longitudinal  deformation  an 
determined  by  experiment. 

Hence,  ax  and  similarly  ay  and  a,,  also,  T*>  Yy  and  r*  may  be  evaluated  as  follows. 


m 


and 


and 


(13J) 


and 

mE 


wherein  G  is  the  modulus  of  tangential  stress  or  shear  and  has  the  value   G     2(jn  +  1)' 

while  m  is  the  "  Poisson  number  "just  given. 

Inserting  the  values  given  by  Eqs.  (18,)  into  Eq.   (13D),  the  fo.lowmg  equat.on  » 
obtained  for  the  actual  work  of  an  isotropic  body : 


.     .     (13K) 


\lso  by  substititingthe  values  from  Eqs.  (13i)  into  Eq.  (13n),  reducing  and  integrat- 
in,  tPS^/AT.,  etc.,  the  following  simple  form  is  obtamed  by  msertmg  the 
value  from  Eq.  (13KJ,  when  temperature  effect  is  negled 

VR^^-      and  similarly       S^Jr  =  ^,  etc.  -     -     (13L) 

OAa 

When  the  abutments  are  immovable  and  no  temperature  effects  exist  then 


£-0       and      £-0,  etc., 


MCT  ab<)ve         fed 
Pm,  or  redundant  Xm:  ^ 


KINETIC  THEORY  OF   ENGINEERING  STRUCTURES 


CHAP.  Ill 


The  value  of  A  in  all  the  Eqs.  (13L)  to  (13N)  is  that  given  by  Eq.  (13K). 

\Yhen  the  abutments  are  immovable  the  last  term  of  Eq.  (13N),  becomes  zero  and 
Castigliano's  law  is  again  established. 

Proceeding  from  Eq.  (13o),  and  writing  same  for  two  related  conditions  of  unit 
loading,  it  is  easily  shown  that  Maxwell's  law  likewise  applies  here. 


ART.  14.     WORK    OF   DEFORMATION   DUE    TO   SHEARING   STRESS 

For  a  beam  of  any  section  and  any  loading  applied  in  the  vertical  plane  YY,  Figs. 

and  14n,  the  resultant  of  the  external  forces  on  one  side  of  any  section  A  A  may  be 
represented  by  a  force  R  which  may  in  turn  be  resolved  into  a  normal  force  N  and  a 
tangential  force  Q  acting  at  the  point  of  application  of  R  on  the  section  A  A. 


T,, 


FIG.  14A. 


Let  M  be  the  static  moment  of  the  normal  force  N  about  the  Z  axis  and  Iz  the  moment 
of  inertia  of  the  section  about  this  same  axis. 

Then  the  unit  stress/  at  any  point  of  the  section  is  by  Navier's  law: 

N     My 
f=J+~r (14.A) 

•*  Z 

If  now  this  stress  undergoes  a  small  change  df  due  to  the  differential  change  in  R 
for  a  neighboring  section,  distant  dx  from  the  first,  then  the  shear  on  the  area  2zidx 
becomes 


2TJ*ldx=f(f+df)dF-ffdF=JdfdF. 


where  rx  is  the  unit  shear  on  the  differential  area  d-F. 

Assuming  N  constant  in  Eq.  (HA)  ,  and  treating  M  as  a  variable,  then  by  differentiation 


(14c) 


AKT.  14     THEOREMS,  LAWS,  AND  FORMULAE  FOR  ISOTROPIC  SOLIDS  39 

Also,  since  the  shear  is  the  differential  of  the  bending  moment,  then 

dM=Qdx (14D) 

Combining  Eqs.  (14B),  (14c)  and  (14o),  then 

CydMdF    Qdx  C  , 

1-xz-idx=  I  I  ydF. 04E) 

J        Lz  lz  J 

and  since  J  ydF  is  the  static  moment  Ms  of  the  cross-section,  this  integration  may  be 
considered  performed  to  obtain 


Return  now  to  Fig.  14s,  take  any  section  pq  perpendicular  to  the  ?/  axis  and  let  Q 
produce  a  unit  shear  r  in  any  point  i  of  this  section.  The  line  of  this  shear  intersects  the 
?/  axis  in  a  point  H,  which  is  determined  by  the  tangent  pH.  This  is  a  pure  assumption, 
but  from  the  nature  of  the  case  it  is  the  most  rational  assumption  to  make. 

The  shear  r  may  now  be  resolved  into  components  TU  and  TZ,  while  rx,  for  this  sectional 
plane,  must  be  zero.  However,  ry  now  has  the  same  value  as  previously  found  for  -x, 
and  hence 

*-!&-£>'• <-> 

Again,  from  Fig.  MB,  tan0  =  —  =  -  from  which  rg  =  -ry.     But  tan  d  =  — ,  therefore, 

O7  i  —  f)  P  P 


/  *   \  j.          n  ft  A     \ 

Ta  =  Tj,(  — itan  0 (14H) 

\zi/ 

For  0=0,  T2=0,  which  is  the  case  for  any  surface  point,  the  tangent  to  which  is  parallel 
to  the  y  axis. 

From  Eqs.  (14r)  and  (14o)  it  follows  that  for  a  given  loading  and  section,  the  shear- 
ing stresses  depend  only  on  M8  and  attain  a  maximum  when  the  section  coincides  with 
a  gravity  axis  parallel  to  the  neutral  axis.  In  practical  cases  it  is  usually  sufficient  to 
consider  the  shearing  stress  of  a  section  as  extending  only  over  a  unit  length  of  the 
beam  as  given  by  Eq.  (14F).  The  actual  work  produced  by  shearing  stress  alone  is 
then  found  from  Eq.  (13K)  as 


By  substituting  the  value  of  r,  from  Eq.  (14n),  into  Eq.    (14i),  the  latter  becomes 


/f 


(14K) 


40 


KINETIC  THEORY  OF   ENGINEERING  STRUCTURES  CHAP.  Ill 


and  by  substituting  for  Ty  its  value  from  Eq.  (14c),  then  Eq.  (14K)  may  be  solved  for  any 
special  section.     This  gives  in  general  for  any  section 


A=> 


and 


and 


(14L) 


In  Eqs.  (14L)  ft  is  an  involved  function  of  the  shape  of  the  section,  called  by  German 
authors  the  "  distribution  number  "  of  the  section,  or  "  coefficient  of  shearing  strain." 

This  number  @  is  1.2  for  any  solid  square  or  rectangular  section,  and  1.111  for  a  solid 
circular  or  elliptic  section.  Other  values  will  be  given  after  illustrating  the  computation 
of  this  number  by  the  use  of  Eq.  (14L). 

bh3 
For  a  rectangle  of  height  h  and  breadth  b  =  2zi,  then  1,.  =  -^;   F  =  bh;    and  dF  =  bdy, 

\m 

hence, 

Q 


and 


For  any  I  section,    Professor  Mehrtens  finds  the  general  formula  for  /?  as  follows: 
{3  =  M\T2dF  =  j2\-£-(8b3—l2b2t  —  t3  +  6bt2)+—(-^—-\ —  -+t'2a)    ,      .     .     (14ivi) 

wherein  i= — - —  and  the  special  dimensions  here  used  are  shown  in  Fig.  14c. 

Professor  L.  von  Tetmajer,  in  his  "  Elastizi- 
taets  und  Festigkeitslehre,"  1905,  p.  49,  gives  a 
table  of  values  for  /?  for  I  beams  ranging  from 
3i"-4  Ibs.  to  19£"-95  Ibs., 'for  which  /3=2.39 
to  2.03  respectively.  Also  for  riveted  girders 
made  of  f"  webs,  4  L's  3&"X3&"Xf"  each, 
and  2  plates  8f"  xf"  on  each  chord.  Then  for 


_!..  __, , 


-Y 


2.71 


23$" 

2.49 


27  A" 
2.35 


The  value  £  is  independent  of  the  unit  of  length  and  is,  therefore,  the  same    for 
metric  and  U.  S.  measures. 


ART.  1.)      THEOREMS,  LAWS,  AND  FORMULAE  FOR  ISOTROPIC  SOLIDS  41 

ART.   16.     WORK    OF    DEFORMATION    FOR    ANY    INDETERMINATE    STRAIGHT 

BEAM 

Direct  ^tress.     For  axial  or  direct  stress  only,  Eq.  (13K)  gives 

(ISA) 
Also,  by  observing  that  ax=fx/E,  then  Eq.  (13n)  becomes  for  direct  stress  alone, 

-    ........  <15B> 


Now   (  fxdF=N,  the  normal  direct  stress,  and,  as  dV  =dFdx,  then  from   Eq.  (15A) 

C&HV-  CNf*dx-  CNf*Fdx_  CN*dx 
J  2Ea[  'I  ~2E~~J     2EF       J  2EF  ' 

and  similarly  from  Eq.  (15B) 

ffxdfxdV_  CN-dfxdx_^  (N-dfxFdx_  fNZNdx  ( 

J    E-dXa       J    EdXa      J    EF3Xa     J  EFdXa 


The    temperature    effects  in  Eqs.   (15A)   and   (15B)   will  now  be  determined  on  the 
supposition  that  the  effect  is  not  uniform,  but  as  shown  on  Fig.  14A,  where 

fo=the  change  in  temperature  from  normal  at  the  gravity  axis  of  the  section; 
J£=the  difference  in  temperature  of  the  two  extreme  fibers; 

h  =  height  of  the  section; 

t/=any  ordinate; 

t=the  temperature  above  normal,  of  any  point  of  the  section  then 


(15B) 


Substituting  this  value  for  t  into  the  temperature  element  of  Eq.  (15A),  the  latter 
becomes 


But    ffxdF=N  and    r/x?/dF=M=moment  of  resistance  of  the  section,  therefore, 

dx  .........     (15F) 


By  differentiating  Eq.  (15?),  and  dividing  through  by  3Xa,  the  value  of  the  tem- 
perature element  for  Eq.  (15u),  is  obtained  thus: 


n  .  . 

(15G) 


42  KINETIC  THEORY  OF   ENGINEERING  STRUCTURES          CHAP,  in 

The  bending  moment.     From  Eq.   (HA),    fx=My/Ig    and    Iz==fy2dF.     With    these 
values  and  neglecting  the  temperature  element,  Eq.  (!OA)  becomes 


By  differentiating  Eq.  (Ion)  and  dividing  through  by  "dXa  then 

'M 


The  general  work  equation,  including  all  effects,  may  now  be  written  out  by  collect- 
ing the  several  Eqs.  (14L),  (15A),  (15c),  (15n),  and  (lor)  into  the  following,  represent- 
ing the  total  actual  work  for  any  isotropic  body  by 

r  r      M 

Similarly  the  following  differential  expression  of  Eq.  (!OK)  is  obtained  from  Eqs. 
(14L),  (15s),  (15D),  (15.i)  and  (15o)  or  directly  by  differentiating  Eq.  (!OK)  and  divid- 
ing through  by  ~dXa,  thus: 


-   •    (loL) 

Equation  (15L)  being  true  for  any  force  Xa  is,  of  course,  true  for  a  force  Pm,  hence 
by  substitution  of  the  latter  value  and  then  inserting  the  value  3A/3PTO.  thus  obtained 
from  Eq.  (15L)  into  Eq.  (13N)  the  following  equation  for  elastic  deflection  is  obtained: 


c  Q   ^Q  ,      C  t  ^N 

j  GFZP^dx+J  £t°3p- 


3M 


The  last  three  equations  (15K),  (15i.)  and  (15M)  are  the  three  fundamentals  from  which 
all  cases  of  redundancy  for  isotropic  bodies  can  be  solved.  There  are  always  as  many 
of  these  equations  as  there  are  redundant  conditions  X. 

In  each  of  these  the  first  term  expresses  the  effect  due  to  direct  or  normal  stress;  the 
second  term  that  due  to  pure  bending;  the  third  term  that  due  to  shearing  stress;  the  fourth 
term  gives  the  effect  due  to  a  uniform  rise  in  temperature  to;  and  the  fifth  term  that  due  to  a 
difference  At  in  the  temperature  of  opposite  extreme  fibers. 


ART.  16     THEOREMS,  LAWS,  AND  FORMULA  FOR  ISOTROPIC  SOLIDS  43 

The  object  in  presenting  these  rather  long  fundamental  equations  is  not  to  make 
the  subject  appear  complicated,  but  rather  with    a  view  to  showing  once  for  all 
combined  effect  from  all  causes,  thus  permitting  the  easy  choice  in  combining  any  effects 
or  in  omitting  such  as  may  seem  negligible  in  any  specific  problem. 

Equation  (15K)  may  also  be  derived  from  Eq.  (5n)  for  virtual  work  and  offers  very 
useful  applications. 

Thus  assuming   a  column  subjected    to  a  thrust  Nttt  and  a  bending  moment   Mm 
the  virtual  work  on  the  column  from  Eq.  (OH)  becomes 


where  Arrt  and  Ma  are  due  to  the  usual  conventional  loadings.     But 

A^te        nd       .„     MJx 


lioncG 

W        .     »  ClNmNadx    ,     rlMrnMadx 

W  =  \-dm 


EF 


rl  MmMadx 
J0        El 


When  the  virtual  work  becomes  the  actual  work  A  then  Eq.  (15N)  becomes  identical 
with  Eq.  (15K)  term  for  term. 

ART.   16.     WORK   OF    DEFORMATION   DUE   TO    DYNAMIC    IMPACT 

Problems  involving  the  deflection  or  strength  of  a  structure  subjected  to  impact, 
are  frequently  met  with,  and  their  treatment  is  here  considered  as  properly  belonging 
to  the  subject  of  the  present  chapter. 
Let  LO  =  weight  of  a  moving  body; 
H  =  height  of  a  fall; 
v=  velocity  at  instant  of  impact; 
g  =  acceleration  due  to  gravity  ; 
a=any  elastic  displacement  produced  by  the  moving  body  in  some  structure. 

Then  the  work  expended  by  the  moving  mass  is  represented,  either  in  terms  of 
velocity  or  height  of  fall  as  follows: 


wherein  v*/2g  represents  the  velocity  height  or  height  through  which  a  body  falls  in 
acquiring  a  velocity  v.  When  the  body  moves  with  a  velocity  v  along  a  horizontal  path, 
thenv2/2g  will  be  the  height  to  which  the  body  would  raise  itself  in  order  to  expend  its 
energy  and  come  to  a  state  of  rest. 


44  KINETIC  THEORY   OF    ENGINEERING  STRUCTURES          CHAP.  Ill 

The  work  thus  produced  by  kinetic  energy,  when  it  is  instantly  taken  up  by  a 
quiescent  body  or  structure,  is  twice  as  great  as  the  work  of  the  same  moving  body 
gradually  applied,  and  hence  produces  twice  the  internal  work  in  the  body  struck  as 
would  result  from  an  equivalent  static  load. 

Problem  1.  The  weight  w  falls  on  top  of  a  column,  stressing  the  column  normally, 
what  sectional  area  is  required  in  order  that  the  unit  compressive  stress  does  not  exceed  /? 

From  Eq.  (!OK)  the  work  due  to  direct  stress  is  given  by  the  first  term,  where  N 
is  the  total  static  load,  and  to  balance  the  dynamic  energy,  twice  this  amount  is  taken; 
hence 


(16B) 


and  since  o  =  ~  this  gives  for  F, 

m 


Problem  2.  A  weight  w  falls  from  some  height  striking  a  beam  resting  on  two  sup- 
ports. The  weight  strikes  the  center  of  the  beam  with  a  velocity  v;  what  will  be  the 
stress  /  in  the  extreme  fiber  for  a  given  beam  section?  Shear  and  bending  resistances 
are  to  be  considered. 

From  Eq.  (!OK)  the  actual  internal  work  is 


Let  P=an  equivalent  static  load  producing  the  same  stress  /  in  the  given  beam. 
Then,  for  any  point  of  the  beam  of  depth  h  and  span  I, 


P  My     Plh  8/7 

-x;      also      /=--*  =  __      or      P  =  -^-  .....     (16s) 


Hence 

i 


, 

+JM2dx_  P2   rt  2     _  P2/3  _2f2ll 

i_  ~2ET~4EI  Jo  X          WEI~3Eh2 


(16r) 


p 

Also,  for  Q  =—  the  second  term  of  Eq.  (16D)  becomes 


=dx  =          _ 
2GF     4GFjo  SGF     GFlh2  ....... 


2 


The  sum  of  Eqs.  (16p)  and  (16c)  gives  A,    according  to  Eq.  (16D),  and  this  must 
equal  a}V*/2g.     Therefore, 


AKT.  16     THEOREMS,  LAWS,  AND  FORMULA  FOR  ISOTROPIC  SOLIDS  45 


Here  G=modulus  of  shear  =  1  =0.385  E  for  structural  steel  (see  Eqs.  (13j). 


The  coefficient  of  shearing  strain  /?  is  given  by  Eqs.  (14L)  and  (14M). 

By  applying  Eq.  (16n)  to  a  steel  beam  2.5X2.5  inches  by  78  inches  long,  it  is  found 
that  when  shear  is  included,  the  stress  /  is  about  f  per  cent  smaller  than  when  this  term 
is  omitted,  showing  that  the  internal  resistance  due  to  shear  is  very  small  and  usually 
negligible.  '  This  is  also  true  when  computing  deflections. 

'  In  the  same  manner  all  problems  involving  impact  may  be  solved. 

Suppose  a  ship  weighing  w  tons,  and  moving  with  a  velocity  of  v  feet  per  second 
were  to  collide  with  a  fixed  structure,  then  the  work  which  the  ship  is  capable  of  expend- 
in-  is  represented  by  a>v*/2g  ft.tons.  This  work  may  be  expended  in  injuring  the 
structure  or  the  ship  itself,  a  condition  depending  on  the  relative  strength  of  the  two 
bodies,  which  must  be  ascertained  form  the  design  and  construction  of  each. 


CHAPTER  IV 
INFLUENCE    LINES   AND   AREAS   FOR    STATICALLY   DETERMINATE    STRUCTURES 

ART.  17.     INTRODUCTORY 

Professors  MOHR  and  WINKLER,  in  1868,  published  simultaneously  the  first  treatises 
on  influence  lines  describing,  at  that  early  date,  practically  all  the  uses  and  applications 
of  these  lines  known  at  the  present  time.  Professor  J.  Weyrauch,  in  1873,  introduces  the 
name  influence  line  not  used  by  Mohr  and  Winkler  in  their  earlier  work.  Professor  Mohr, 
in  a  series  of  articles  published  from  1870  to  1877,  was  the  first  to  apply  influence  lines 
to  deflections  and  to  redundant  conditions. 

Professor  Geo.  F.  Swain,  in  1887,  gave  the  first  treatise  on  the  subject  in  English. 

An  influence  line  is  the  graphic  representation  of  some  particular  effect  produced, 
at  a  certain  point  of  a  structure,  by  a  single  moving  load  occupying,  successively,  all  possible 
positions  over  the  entire  span.  The  effect  may  be  the  shear,  the  bending  moment  or 
the  deflection  at  any  certain  point  of  the  structure;  it  may  also  represent  any  reaction 
force  or  the  stress  in  any  member.  The  single  moving  load  is  usually  taken  equal  to 
unity,  though  in  certain  special  cases  it  may  be  desirable  to  use  any  load  P. 

An  influence  line  represents  a  certain  effect  for  a  certain  point  or  member  of  a 
structure  and  for  any  position  of  a  moving  load,  while  a  shear  or  moment  diagram 
represents  effects  due  to  a  single  position  of  the  load  or  loads  for  all  points  of  a  structure. 

The  ordinate  to  any  influence  line  is  thus  an  influence  number  or  factor,  usually 
designated  by  TJ  when  stresses  are  dealt  with  and  sometimes  by  o  when  deflections  are 
under  consideration. 

As  a  matter  of  convention,  all  positive  influence  line  ordinates  will  be  laid  off  down- 
ward from  the  axis  of  abscissae. 

A  load  point  is  any  particular  one  of  the  many  possible  positions  of  the  moving 
load. 

A  summation  influence  line  is  one  which  gives  the  total  effect  at  some  particular  point 
due  to  a  train  of  concentrated  loads.  The  actual  loads  are  here  employed  and  each 
influence  ordinate  is  made  to  represent  the  summation  of  influences  of  the  same  kind 
for  a  certain  position  of  the  train  of  loads.  Usually  this  position  is  taken  so  that  the  first 
load  is  over  the  point  for  which  the  influence  line  is  constructed. 

An  influence  area  is  the  area  included  between  the  influence  line,  the  axis  of  abseissce 
and  the  two  end  ordinates. 

The  maximum  effect  is  always  produced  when  the  load  point  coincides  with  the 
maximum  ordinate  of  the  influence  area.  Hence  influence  lines  are  eminently  suited 

46 


ART.  17  INFLUENCE  LINES  AND  AREAS  47 

to  the  solution  of  all  problems  relating  to  position  of  moving  loads  for  maximum  and 
minimum  effects  and  stresses. 

The  load  divide  is  such  a  load  point,  the  ordinates  to  either  side  of  which  have 
opposite  signs.  Hence,  the  influence  ordinate  at  a  load  divide  passes  through  zero  and 
the  influence  line  intersects  the  axis  of  abscissae  at  such  load  divide. 

It  is  assumed  that  if  a  unit  load  produces  some  effect  77  at  a  certain  point  of  a  given 
structure,  then  a  load  P  will  produce  the  effect  PTJ  at  this  same  point,  so  long  as  the  material 
is  not  stressed  beyond  the  elastic  limit.  This  follows  from  the  law  of  proportionality 
stated  at  the  end  of  Art.  4. 

Hence,  the  total  effect  Z  produced  by  a  system  of  moving  loads,  PI,  P2,  etc.,  will 
be  the  sum  of  the  effects  Pi]  of  all  the  loads  and  the  maximum  and  minimum  value  of  Z 
will  be  determined  by  the  position  of  the  moving  loads. 

The  total  effect  is  thus  expressed  by  the  following  equation  when  the  case  of  loading 
is  simultaneous  and  within  working  limits  : 


(17A) 


Therefore,  having  given  the  influence  line  for  a  certain  effect  on  some  structure,  then 
the  total  effect,  due  to  any  system  of  loading,  is  easily  found  by  a  summation  of  the  prod- 
ucts of  loads  into  corresponding  influence  ordinates  or  numbers  for  any  desired  positions 

of  the  loads. 

For  a  uniform  moving  load  p  per  foot  of  length,  Eq.  (!?A)  becomes 


wherein  the  integral  represents  the  area  of  the  influence  polygon  between  the  end  ordi- 
nates of  the  uniform  load. 

The  equation  of  any  influence  line  may  be  written  out  by  expressing  the  desired 
function  for  a  particular  point  in  question  in  terms  of  a  moving  load  unity  acting  at  any 
variable  distance  x  from  one  end  of  the  structure  taken  as  origin. 

Since  influence  lines  represent  all  possible  effects  it  is  readily  seen  that  maximum 
and  minimum  stresses  may  be  found  from  the  same  lines.  Therefore,  only  one  half  of 
a  symmetric  structure  requires  analysis.  The  left  half  is  usually  treated,  as  a  matter 

of  conventional  uniformity. 

Direct  and  indirect  loading.  In  the  above  it  was  assumed  that  the  loads  were  directly 
applied  to  the  beam,  which  is  rather  the  excep- 
tion. Usually  they  are  taken  up  by  the  floor 
system  and  then  transferred  to  the  panel  points 
as  load  concentrations.  The  former  case  of 
loading  will  be  known  as  direct  loading  and  the 
latter  as  indirect  loading. 

The  influence  line  between  two  successive  panel 

points  is  always  a  straight  line,  regardless  of  the  riG  1?A 

system  of  loading  or  the  particular  influence. 

Let  Fig.  17A  represent  two  successive  floor  beams  of  any  truss  and  the  loadP  = 


IS  KINETIC  THEORY  OF  ENGINEERING  STRUCTURES          CHAP.  IV 

is  transferred  to  the  points  a  and  b  by  the  floor  stringer  ab.     The  load  effects  PI  and 
Po,  which  are  carried  to  the  panel  points  a  and  b,  are 


-  and 


Now  the  total  influence  of  Pl  and  P2,  on  the  structure  as  a  whole,  must  be  exactly 
equal  to  that  produced  by  the  resultant  P. 

But  the  influence  line  a'b' '',  extending  over  the  panel  ab,  must  give,  by  Eq.  (!?A), 


,17  x 


from  which 

fd-: 


For  any  influence  line,  >?i,  r)2  and  d  are  constants,  hence  Eq.  (17c)  represents  a 
straight  line,  which  was  to  be  proven. 

ART.   18.     INFLUENCE    LINES   FOR   DIRECT    LOADING 

The  several  influence  lines  for  reactions,  shears,  moments  and  deflections  for  a  beam 
with  direct  loading  will  now  be  given.  They  will  always  be  known  by  the  names  indicated 
in  Fig.  ISA.  Thus  the  A  line  is  the  influence  line  for  the  reaction  A. 

The  influence  line  for  a  reaction  is  of  the  first  importance,  because  the  other  lines 
are  generally  derived  from  the  reaction  influence  line.  This  is  really  seen  from  the 
circumstance  that  for  the  unloaded  portion  of  a  beam  or  truss,  the  shear  is  always 
equal  to  the  end  reaction  and  the  moment  is  equal  to  this  reaction  into  the  distance  from 
the  section  in  question  to  the  end  of  the  beam. 

Hence,  in  drawing  influence  lines  it  is  always  best  to  consider  the  unloaded  portion 
of  the  span  to  the  right  or  left  of  the  section  as  the  case  may  be,  beca-use  the  only  external 
force  on  that  side  of  the  section  will  then  be  the  reaction. 

In  the  following  the  moving  load  will  always  be  assumed  as  coming  on  the  span  from 
the  right  end  and  the  effect  produced  on  the  left  half  of  the  span  only,  is  considered. 

(a)  Reaction  influence  lines  A  and  B.  Using  the  dimensions  indicated  in  Fig.  ISA, 
the  two  reactions  become 

Px'  Px 

A=—f-       and       B=~T  .......... 


For  x'  and  x  variable,  both  expressions  represent  equations  of  straight  lines  which 
are  easily  plotted.  When  P  =  l  and  x'  =1  then  A=l  and  for  x'  =0,  A  =0.  Hence,  the 
A  line  is  drawn  by  laying  off  a  distance  unity  down  from  A  and  joining  this  point  with  B. 
The  reaction  A  for  any  load  P  acting  at  the  load  point  m  is  then  Am  =Prjm. 

In  like  manner  the  B  line  is  found,  and  the  corresponding  reaction  Bm  for  a  load 
P  at  m  becomes  Bm=Pj)'m,  and  i?m  +  fy»'  =  l  for  every  point  of  the  span. 


ART.  IS 


INFLUENCE  LINES  AND  AREAS 


(b)  Shear  influence  lines.     The  shear  Qn  for  a  certain  point  n  to  the  left  of  the  load 
point  m  is  always  equal  to  the  reaction  A.     But  for  a  load  point  to  the  left  of  n  (not 
shown)  the  shear  is  Qn  =  A  —  P  =  —B,  because  P  =A  +B. 

The  influence  line  for  Qn  is  thus  derived  from  the  influence  lines  for  A  and  —B  as 
indicated  in  Fig.  ISA,  and  consists  of  the  polygon  An'n"R. 

The  point  n  becomes  the  load  divide  for  shears  at  n  and  hence  there  will  always  be 
a  maximum  and  a  minimum  value  for  Qn  depending  on  whether  the  positive  or  the 
negative  influence  area  is  fully  loaded.     The  shear  due 
to  any  single  load  P  must  change  sign  in  passing  the 
point  n. 

(c)  Moment    influence    lines.     The    moment   for 
the  point  n  is  now  found  when  the  moving  load  is 
to  the  right  or  left  of  the  section  at  n.     Then  for 


P.I 


•  •m 


x>a,    Mn=Aa;  and  for 


Mn=B(l—a). 


Thus  the  moment  influence  line  is  also  derived 
from  the  reaction  lines  because  the  ordinates  of  the 
latter,  when  multiplied  by  a  or  (I—  a)  as  the  case  may 
be,  give  the  ordinates  to  the  moment  line.  Hence 
the  bounding  lines  of  the  moment  influence  line  are 
easily  found. 

Since  A  and  B  are  both  unity,  then  the  ordinate 
A  A'  --=a  and  the  ordinate  BB'=l—a  and  the  lines 
AB'  and  A'B  inclose  the  required  moment  influence 
line.  Also,  the  ordinate  at  n  is  the  ordinate  of  the 
intersection  n'  between  the  two  bounding  lines.  Hence, 
if  BE'  should  fall  off  the  drawing,  then  the  line  ~AB' 
may  be  drawn  from  A  to  n'  without  finding  B' '. 

In  either  case  the  moment  influence  line  is  then 
the  polygon  An'B  with  all  ordinates  positive  so  long 
as  the  point  n  is  not  outside  the  span,  a  case  which 
will  be  given  later.  The  maximum  ordinate  is  always 
under  n  and  has  the  value  l-a(l  —  a)/l,  which  offers 

still  another  construction  for  this  influence  line.  The  middle  ordinate  of  the  AB' 
\me=±(l-a)  which  furnishes  a  convenient  construction  for  this  line  when  B'  falls  off 
the  drawing. 

It  is  clearly  seen  that  if  a  single  load  is  placed  at  n  over  the  maximum  ordinate, 
then  a  maximum  moment  is  produced. 

(d)  Deflection  influence  lines.  The  deflection  for  a  point  n,  produced  by  a  single 
load  at  any  point  m,  is  given  in  terms  of  the  moment  of  inertia  of  the  beam  section  and 
the  modulus  of  elasticity,  as  follows: 


FIG.  ISA 


QEIl 


-* 


a*)  P. 


(18s) 


50 


KINETIC  THEORY  OF  ENGINEERING  STRUCTURES          CHAP.  IV 


This  is  a  cubic  equation  and  when  plotted  for  all  values  of  x  and  making  P  =  l,  the 
influence  line  for  deflections  is  obtained. 

In  general,  any  deflection  polygon,  drawn  for  a  load  unity  acting  on  a  fixed  point 
n  of  any  structure,  is  the  deflection  influence  line  for  the  point  n  of  that  structure.  This 
follows  from  Maxwell's  law. 

As  the  subject  of  deflection  influence  lines  is  fully  treated  in  Chapter  VIII,  no 
further  consideration  is  given  to  it  now. 


B 


ART.  19.     INFLUENCE   LINES   FOR   INDIRECT   LOADING 

In  the  present  case  the  loads  are  applied  to  a  stringer  and  floor-beam  system  and  are 
thus  transferred  to  the  beam  at  certain  panel  points  as  load  concentrations.  The 

beam  is  then  indirectly  loaded  and  since  an  influence 
line  was  shown  to  be  a  straight  line  between  panel 
points  the  present  case  will  require  some  slight 
modification  of  the  influence  lines  just  found  for 
direct  loading.  See  Fig.  19A. 

However,  all  that  was  said  for  cases  of  direct 
loading  applies  here  and  the  modifications  made 
necessary  by  the  indirect  loading  occur  only  in  the 
panel  wherein  the  point  n  is  located.  As  before,  n  is 
the  point  for  which  the  influence  line  is  drawn  and  m 
is  any  one  of  the  possible  load  points  for  the  moving 
loadP  =  l. 

(a)  Reaction    influence    lines    remain    the   same 
whether  the  loading  is  direct  or  indirect,  since  the  point 
for  which  the  influence  is  sought  is  alwavs  at  A  or  B, 

•-1  •/ 

which  are  also  panel  points. 

Furthermore,  the  reaction  influence  lines  being 
straight  over  the  length  of  the  span  will  always  be 
straight  between  successive  panel  points. 

(b)  Shear  influence  lines.    The  Q  line,  outside  of 
the  panel  ce  and  containing  the  point  n,  remains  the 
same  as  for  direct  loading.      But   since  the  influence 
line  within  a  panel  must  be  a  straight  line,  therefore, 
the  points  c'  and  e'  must  determine  the  influence  line 
for  the   panel   ce   regardless   of   the   location   of   the 

point  n  so  long  as  this  point  is  within  the  panel  ce. 

Hence,  the  polygon  Ac'e'B  is  the  shear  influence  line  for  the  panel  ~ce  and  the  point 
i  is  the  load  divide  for  this  panel.  The  limiting  values  for  shear  are  thus  the  same  for 
any  point  n  of  the  same  panel. 

(c)  Moment  influence  lines.  Here  again  the  influence  line  within  the  panel  ~ce  is 
all  that  requires  modification  in  the  case  of  indirect  loading. 


FIG.  19x. 


ART.  20 


INFLUENCE  LINES  AND  AREAS 


51 


Hence  for  the  same  reasons  just  given  to  determine  the  modified  Q  line,  the  M  line 
for  the  point  n  is  now  the  polygon  Ac'e'B,  instead  of  the  triangle  An'B  for  direct  loading. 
The  point  n  is  the  center  of  moments. 


ART.  20.     THE    LOAD   DIVIDE    FOR   A   TRUSS 

In  constructing  influence  lines  for  truss  web  members  it  frequently  happens  that 
one  or  the  other  end  ordinate  falls  outside  the  limits  of  the  drawing. 

There  is  a  very  simple  way  of  locating  the  load  divide  i  and  thus  facilitating  the 
construction  of  any  influence  line.  The  method  is  given  first  and  the  proof  follows: 

Let  Fig.  20A  represent  a  truss  arranged  for  bottom  chord  loading,  but  no  loads 
are  shown.  Required  to  find  the  load  divide  i  for  the  panel  fg  necessary  to  determine 
the  stress  in  the  diagonal  eg  cut  by  the  section  tt. 


FIG.  20A. 

Construction.  Prolong  the  unloaded  chord  member  ek  in  the  panel  fg  until  it  inter- 
sects the  verticals  through  the  end  reactions  in  a  and  6.  Then  the  intersection  i  of 
the  two  lines  af  and  bg  will  be  the  required  load  divide. 

Proof.  Suppose  the  unit  moving  load  is  now  located  in  the  vertical  through  i. 
Then  if  i  is  the  desired  load  divide,  the  unit  load  for  the  load  point  ii  will  produce  zero 
stress  in  the  member  eg  and  the  influence  line  ordinate  for  the  point  i  must  be  zero. 

Let  F  and  G  be  the  panel  concentrations  in  the  points  /  and  g  due  to  the  load  P  =  1 
at  i. 

Then  the  polygon  abfg  may  be  regarded  as  the  equilibrium  polygon  for  the  forces 
A,  B,  F  and  G.  Also,  the  resultant  R  of  all  forces  on  one  side  of  the  section  tt  must  pass 
through  the  intersection  n  of  the  two  included  sides  of  the  equilibrium  polygon. 

But,  the  point  n  is  the  intersection  of  the  two  chords  fg  and  ek,  which  point  is  also 
the  center  of  moments  for  the  diagonal  eg. 

Hence  the  moment  of  this  resultant  R  about  n  must  be  zero  and  cannot  produce  stress 
in  the  member  eg,  thus  proving  that  the  load  P  =  1  is  actually  located  in  the  load  divide. 

When  the  top  chord  is  the  loaded  chord,  a  similar  construction  furnishes  the  point 
i',  as  the  load  divide  for  the  member  eg,  according  to  the  same  proof  above  given. 

This  construction  applies  to  any  structure  even  when  one  or  both  chords  are  straight. 

When  the  center  of  moments  n  falls  inside  the  span  then  there  is  no  load  divide  and 
all  loads  on  the  entire  span  will  produce  the  same  kind  of  stress  in  any  particular  web 
member. 


52  KINETIC  THEORY  OF  ENGINEERING  STRUCTURES          CHAP.  IV 


ART.  21.     STRESS   INFLUENCE    LINES   FOR   TRUSS   MEMBERS 

According  to  Professor  Ritter's  well-known  method  of    sections  given    in  Art.  58, 
the  stress  in  any  member  of  a  determinate  truss  may  be  expressed  thus  : 


(21  A) 


where  M  is  the  moment  of  all  the  external  forces,  including  the  reactions,  on  one  side 
of  the  section  and  r  is  the  lever  arm,  of  the  member  in  question,  measured  from  a  center 
of  moments  determined  by  the  intersection  of  the  two  other  members  cut  by  the  section. 

Hence,  the  influence  line  for  the  stress  in  an}^  member  is  the  influence  line  for  the 
corresponding  moment  for  this  member,  the  ordinates  of  which  are  divided  by  r. 

For  any  moment  influence  line,  the  ordinate  at  any  point  represents  the  moment  due 
to  a  unit  load  at  that  point  and,  therefore,  from  Eq.  (2lA),  i)=M=rS. 

But  the  end  ordinates  for  a  moment  influence  line  were  formerly  found  to  be  a  and 
a',  therefore,  the  end  ordinates  for  a  stress  influence  line  must  be  1/r  times  as  great, 
whence 


o' 


AA'=-,      and      BB"=- (21u) 

r'  r 

Now  l-a/r=$0=the  stress  in  a  member  S  due  to  a  reaction  unity  at  A,  when  the 
load  producing  this  reaction  is  to  the  right  of  the  section  through  S. 

Also  I  •  a' /r  =${,=  the  stress  in  this  same  member  S  due  to  a  reaction  unity  at  B 
when  the  load  producing  this  reaction  is  to  the  left  of  the  section. 

Hence,  the  end  ordinates  for  any  stress  influence  line  are  equal  to  the  unit  stresses 
Sa  and  Sb  produced  in  the  member  by  reactions  unity  at  A  and  B  respectively. 

Since  these  stresses  are  readily  determined  either  by  Ritter's  method  or  by  a  Max- 
well diagram,  all  stress  influence  lines  for  the  members  in  any  determinate  truss  are 
easily  found. 

In  Fig.  2lA,  the  stress  influence  lines  are  drawn  for  a  general  case  of  truss  design. 
The  top  chord  is  the  loaded  chord  and  the  influence  lines  are  drawn  for  the  three  mem- 
bers L,  U  and  D  of  the  panel  cut  by  the  section  it. 

£te=the  stress  in  the  member  L  for  a  reaction  unity  at  A.  This  is  evaluated  by 
Ritter's  method  in  terms  of  the  lever  arms  ai  and  rt. 

Sib=ihe  stress  in  the  member  L  for  a  reaction  unity  at  B  and  is  expressed  in  terms 
of  the  lever  arms  a'i  and  rt. 

The  stress  influence  line  for  the  bottom  chord  L,  is  drawn  in  accordance  with  the 
method  illustrated  in  Fig.  19A,  observing,  however,  that  the  end  ordinates  now  become 
the  stresses  Sta  and  Sn,  respectively. 

The  center  of  moments  for  this  member  being  at  E,  which  is  vertically  over  E', 
the  stress  influence  line  for  the  member  L  becomes  a  triangle  AE'B,  determined  by  the 
end  ordinates  A  A'  =£to  and  BB'  =£#. 

The  influence  line  for  the  top  chord,  J7  is  drawn  in  exactly  the  same  manner,  but  in 
this  instance  the  resulting  triangle  AF'B  does  not  offer  a  straight  line  over  the  panel 


ART.  21 


INFLUENCE   LINES  AND   AREAS 


53 


Here  F  is 


PJG  and  hence  the  line  E'G'  must  be  drawn  to  complete  the  influence  line, 
the  center  of  moments  for  the  member  U. 

The  influence  line  for  the  diagonal  D  is  similarly  drawn  and  the  load  divide  %' ',  found 
in  the  lower  diagram,  is  seen  to  coincide  with  the  point  i  found  on  the  truss  diagram 
by  the  method  previously  given.  Also  the  intersection  0',  between  the  limiting  rays 

is  verticallv  under  the   center   of  moments  0  for  the 


AE'   and  G'B  of  the  D 
member  D. 


line. 


ISTRESS  INFLUENCE  LINE  FOR  L. 


STRESS  |NFJLUE!NCE  LINE  FOR.  u." 

'  '  " 


STRESS  INKLUEJNCE  LINE  FOR  o 


It  should  be  observed  that  in  all  three  influence  lines,  Fig.  21  A,  the  limiting  rays  intersect 
on  the  vertical  through  the  center  of  moments  for  the  particular  member.  This  then  serves 
as  a  check  on  the  diagram. 

Regarding  the  sign  of  the  influence  area  the  following  rule  should  be  observed: 
When  the  center  of  moments  is  located  between  the  supports  A  and  B  then  all  ordinates  of 
that  particular  stress  influence  line  will  be  of  the  same  sign;  when  the  center  of  moments 
is  off  the  span,  then  there  exists  a  load  divide  and  there  will  be  positive  and  negative  influence 
areas  giving  rise  to  stresses  of  opposite  signs. 


54  KINETIC  THEORY  OF  ENGINEERING  STRUCTURES         CHAP,  iv 

The  criterion  for  position  of  a  moving  load  for  maximum  and  minimum  stresses  in 
any  member  is  thus  clearly  shown  by  influence  lines  and  the  maximum  effect  of  any  load 
is  always  produced  when  that  load  is  situated  over  the  maximum  influence  ordinate. 
This  subject  is  more  fully  treated  in  another  article.. 

It  is  sometimes  necessary  to  construct  influence  lines  without  the  divisor  r,  and 
dividing  the  final  stress  thus  found  by  r.  Either  method  is  employed  as  circumstances 
may  demand,  though  it  is  always  better  to  avoid  the  divisor  and  follow  the  method  given 
in  Fig.  21  A,  unless  there  is  some  very  good  reason  for  doing  otherwise. 

Attention  is  also  called  to  the  fact  that  moment  and  shear  influence  lines  are  not 
required  when  the  stress  influence  lines  are  used.  The  former  were  given  to  render  the 
treatment  complete  and  may  be  employed  to  find  stresses,  but  it  is  usually  preferable 
to  draw  the  stress  influence  lines  directly.  The  final  result  of  an  analysis  will  be  more 
satisfactory  when  this  is  done. 

Influence  lines  are  seldom  used  for  finding  dead  load  stresses,  though  such  a  procedure 
may  sometimes  be  warranted,  and  then  the  following  points  must  be  observed  :  For 
direct  loading  and  for  beams,  it  makes  no  difference  whether  the  influence  lines  be 
used  for  dead  or  live  loads,  but  in  dealing  with  framed  structures  an  error  might  be 
committed  because  then  the  influence  lines  are  always  drawn  for  a  certain  loaded 
chord  while  the  dead  loads  act  along  both  chords. 

This  is  easily  remedied  by  drawing  the  influence  lines  for  both  chords  loaded  and 
applying  the  dead  loads  to  the  proper  lines.  These  influence  lines  are  identical  except 
in  the  panel  containing  the  member  in  question.  Hence  by  observing  this  circum- 
stance, dead  load  stresses  may  be  found  from  the  same  influence  lines  without  difficulty. 

ART.  22.     REACTION   SUMMATION   INFLUENCE   LINES 

Most  summation  influence  lines  become  rather  complicated  and,  therefore,  their 
use  is  practically  restricted  to  a  few  cases  for  which  the  construction  is  simple. 

In  general,  any  ordinate  to  a  summation  influence  line  is  expressed  by  Eq.  (17A),  as 


and  it  is  readily  seen  that  when  P  and  ij  are  both  variables,  such  a  line  would  ordinarily 
require  much  labor  for  its  determination. 

However,  the  summation  influence  line  for  an  end  reaction  of  a  simple  truss  is 
very  easily  constructed  and  serves  a  most  valuable  purpose  in  finding  the  end  shears 
for  a  system  of  moving  loads.  This  influence  line  will  be  called  simply  the  sum  A  line 
to  distinguish  it  from  the  ordinary  A  line  in  Figs.  ISA  and  19A. 

The  general  usefulness,  of  the  sum  A  line  originated  by  Professor  Winkler,  will  be  shown 
later;  suffice  it  to  say  here  that  it  affords  a  ready  means  of  finding  stresses  in  the  web 
members  of  any  truss,  because,  as  previously  shown,  the  shears  and  moments  for  any 
point  of  a  truss  are  easily  found  when  the  end  reaction  for  the  particular  loading  is  known. 

The  sum  A  line  for  a  system  of  concentrated  wheel  loads    will  now  be  demonstrated 
using  but  five  loads  for  simplicity,  though  the  method  applies  to  any  number  of  loads. 


ART.  22 


INFLUENCE  LINES  AND  AREAS 


55 


Given  the  train  of  moving  loads  PI,  P2,  etc.,  in  kips  of  1000  Ibs.,  and  spaced  as 
shown  in  Fig.  22A,  to  construct  the  sum  A  line  for  a  span  of  25  ft. 

Since  a  reaction  influence  line  is  the  same  both  for  direct  and  indirect  loading,  the 
sum  A  line  is  independent  of  the  number  and  location  of  panel  points. 

For  standard  position  of  loads  on  a  truss,  the  train  is  always  assumed  as  coming 
on  the 'span  from  the  right  end  and  moving  toward  the  left,  but  the  above  loads  are 
placed  in  exactly  the  reverse  order,  with  the  first  load  PI  over  the  support  B.  The 
reason  for  this  will  appear  later. 

The  loads  are  applied  consecutively  from  PI  up  on  the  vertical  through  A,  using 
any  convenient  scale. 


r 


POSITJON  OF  LOADS  roft  CONST *GcT  ION  or  SUM  A  unr 


B 


FIG.  22A. 


The  several  rays  drawn  from  B  to  the  respective  load  points  c,  d,  e,  etc.,  form 
a  force  polygon  with  pole  distance  H  =1.  The  sum  A  line  is  then  drawn  as  the  equilibrium 
polygon  for  this  force  polygon  and  the  loads  P,  by  making  1  -2  \\Bd,  2  -3  \\Bc,  3  -4  \\Bf, 
etc. 

It  will  now  be  shown  that  the  equilibrium  polygon  so  obtained  is  really  the  sum- 
mation influence  line  for  the  end  reaction  A . 

Referring  again  to  Fig.  22A,  the  line  Be  is  easily  recognized  as  the  influence  line 
for  A  due  to  a  moving  load  PI.  Alsp^the  influence  line  for  A  when  the  moving  load  is 
P2,  may  be  represented  by  the  line  Bd  when  Be  is  the  axis  of  the  absciss®.  Similarly 
~Be  represents  the  A  line  for  P3,  and  so  on  for  any  other  loads.  The  summation  of 
all  odinrates  should  then  represent  the  desired  summation  influence  line. 

For  the  train  coming  on  the  span  from  right  to  left,  the  ordinate  TJ,  under  the 
forward  load  PI,  always  represents  the  reaction  A.  Thus  when  the  train  has  advanced 


56  KINETIC  THEORY  OF  ENGINEERING  STRUCTURES         CHAP.  IV 

to  the  point  n,  Fig.  22A,  the  ordinate  v?  represents  the  sum  of  the  ordinates  t,  and 
T}=An=U+t3+t2+ti\vhkb.  is  readily  seen  by  inspection.  The  ordinates  t  are  the  con- 
tributions to  A  from  the  several  loads  on  the  span.  It  is  now  seen  why  the  train 
was  first  placed  in  reverse  position  for  the  construction  of  the  equilibrium 
polygon. 

Hence  the  sum  A  line  for  direct  or  indirect  loading  is  an  equilibrium  polygon  drawn 
for  a  pole  distance  equal  to  the  length  of  span  and  any  system  of  train  loads  placed  in 
reverse  order  on  the  span  with  the  forward  load  at  B. 

The  reaction  A  for  any  position  of  loading  is  then  the  ordinate  to  the  sum  A  line 
measured  under  the  forward  load  when  the  train  approaches  from  right  to  left.  The 
greatest  possible  reaction  A  will  always  be  the  end  ordinate  A5,  Fig.  22A. 

When  extraordinary  accuracy  is  required  the  end  ordinate  A 5  can  be  readily  checked 
by  computation. 

The  sum  A  line  for  uniformly  distributed  loads  may  be  constructed  in  precisely  the 
same  manner  as  above  illustrated  for  concentrated  loads,  merely  by  laying  off  the  load 
line  Ag  to  represent  the  total  load  on  the  span  =pl  and  dividing  this  into  some  con- 
venient number  of  equal  parts.  The  greater  this  number  the  more  accurate  will  be  the 
result,  and  the  polygon  finally  becomes  a  parabola. 

The  application  of  the  sum  A  line,  to  finding  the  shear  at  any  point  of  a  beam  or  truss, 
will  now  be  presented. 

The  shear  at  any  point  n  of  a  beam,  for  a  case  of  direct  loading,  is  equal  to  the 
reaction  A  minus  the  loads  between  A  and  n.  Hence,  the  sum  A  line  gives  a  complete 
solution  for  this  case.  The  subtraction  of  any  loads  to  the  left  of  n  can  be  performed 
graphically  on  the  diagram,  Fig.  22 A,  by  taking^  off  the  proper  ordinate  less  the  loads 
between  A  and  n. 

For  indirect  loading,  the  shear  at  n  is  equal  to  the  end  reaction  A,  provided  the 
forward  load  is  exactly  over  the  panel  point  n.  But  when  the  load  extends  over  into 
the  panel,  then  the  shear  is  equal  to  the  end  reaction  A,  minus  the  panel  reaction  a  at 
the  left  hand  pin  point  of  the  loaded  panel.  See  Fig.  22ra. 

The  values  of  o  for  all  possible  positions  of  loads  in  a  single  panel  may  be  found 
by  drawing  a  sum  a  line  noe  for  one  panel,  using  as  many  of  the  loads  from  the  forward 
end  of  the  train  as  may  be  placed  into  one  panel.  If  this  auxiliary  sum  a  line  is  drawn 
on  tracing  paper,  it  will  also  serve_  a  ready  means  for  finding  the  position  of  the  train 
for  maximum  shear  in  any  panel  mn. 

The  method  of  finding  the  sum  a  line  is  exactly  the  same  as  for  the  sum  A  line  only 
the  pole  distance  for  the  former  is  the  panel  length  d. 

When  the  forward  wheel  of  the  train  is  at  n,  then  the  shear  is  the  ordinate  rjn. 
When  the  forward  wheel  is  at  r  the  shear  in  the  panel  mn  is  Qr=rc—re=ec  where  re  =  -na  — 
the  panel  reaction  a  for  loads  PI  and  P2  in  the  panel  and  load  P3  at  n.  Similarlv  when 
the  forward  wheel  is  at  s  and  the  second  wheel  is  at  n  the  shear  Q8  =s/"— os. 

The  position  of  the  train  for  max.Q  in  the  panel  mn  is  then  easily  found  by  selecting 
such  a  point  s  for  which  the  ordinate  O/=T?  max.  This  maximum  jj  is  easily  found  with 
a  pair  of  dividers  and  will  always  occur  when  some  one  of  the  forward  wheels  is  at  the 
panel  n. 


INFLUENCE  LINES   AND  AREAS 


57 


If  the  sum  a  line  is  drawn  on  tracing  cloth,  it  may  be  superimposed  on  any  panel 
of  the  sum  A  line  and  TJ  max.  can  be  quickly  determined. 

It  sometimes  happens  that  two  positions  of  the  train  give  the  same  maximum  shear, 
in  which  case  either  may  be  used.  In  Fig.  22e  this  would  be  true  if  oe\\cf  then  ~ec=of 
and  both  Qr  and  Qs  would  thus  be  equal. 


Load  potitiori  for  Qm«<.  in  panel  m-n 


FIG.  22s. 


ART.   23.     POSITIONS   OF    A    MOVING   TRAIN   FOR   MAXIMUM   AND    MINIMUM 

MOMENTS 

(a)  Point  of  greatest  moment  for  direct  loading.  Given  a  span  of  certain  length 
I,  loaded  with  a  train  of  loads,  PI,  Pa,  PS,  etc.,  to  find  the  point  n  which  is  subjected 
to  the  greatest  bending  moment.  It  is  assumed  that  all  loads  remain  on  the  span. 
See  Fig.  23 A. 

Let  R  =  resultant  of  all  loads  P  on  the  span. 
RI  =  resultant  of  all  loads  to  the  left  of  n. 
R2  =  resultant  of  all  loads  to  the  right  of  n. 
A  and  B  are  the  end  reactions  for  R. 
n  =the  point  of  maximum  moments  to  be  found. 


Then  the  moment  about  n  is 


Rx 


(22A) 


58  KINETIC  THEORY   OF  ENGINEERING  STRUCTURES          CHAP.  IV 

By  differentiating  and  equating  to  zero,  the  abscissa  for  the  point  of  maximum 
moments  is  found  thus  : 

dM    R  n  a      I  r9.,  x 

-=-r(l-2x-a)  =0       or       x+-=-^  .......     UOB) 

ax      t  z      z 


Eq.  (23B)  signifies  that  the  maximum  moment  under  any  system  of  loads  occurs  for 
a  point  n  when  the  center  of  gravity  of  the  system  and  the  point  n  are  equidistant  from 
the  center  of  the  beam. 

Introducing  the  value  of  x  from  Eq.  (23  B)  into  Eq.  (23A)  the  value  of  max.  M  is 
obtained  as 

max.M^R^-R^.       .     .     ........     (23c) 

l> 

In  any  special  case  it  is  necessary  to  find  the  particular  load  Pn  which  must  act 
at  the  point  n  to  produce  a  real  maximum.  The  moment  influence  line  clearly  indicates 
that  one  of  the  loads  must  fall  at  n  to  obtain  the  maximum  moment. 


FIG.  23A. 

Usually  the  point  n  falls  very  near  the  center  of  the  beam  and  in  most  practical 
problems  it  will  suffice  to  place  that  load  at  n  which  falls  nearest  the  center  of  gravity 
of  the  sj'stem  of  loads. 

When  dealing  with  floor  stringers  it  frequently  happens  that  the  stringer  is  long 
enough  to  carry  three  wheel  loads  but  that  the  max.  M  so  found  is  less  than  when  only 
two  of  the  loads  are  on  the  stringer.  A  few  of  these' special  cases  are  here  given. 

Case  I.  When  there  is  only  one  load  P  on  the  span,  then  x=l/2  and  a=0. 
Hence  Eq.  (23c)  gives 

M      Pl 

max.M=— (23D) 

Case  H.  When  there  are  two  equal  loads  P  on  the  span,  distant  e  from  each  other, 
then  R=2P  and  |=|,  hence  z=--|=_- 1  which  gives,  from  Eq.  (23c), 


™     2P/Z 
max.  M  =—:-[  — 


Equating  Eqs.  (23o)  and  (23c)  to  find  when  one  or  two  loads  give  equal  maxima, 
this  results  in  the  condition  e=0.5858Z.  Hence,  when  e>0.5858Z,  then  one  load  gives 
a  greater  moment  than  two  loads,  all  loads  being  of  same  magnitude. 


ART.  23 


INFLUENCE  LINES   AND   AREAS 


59 


Case    III.      When  there  are  three  equal  loads  P  on  the  span,  distant  e  from  each 
other,  then  R  =3P  and  x  =1/2  making  a  =-0.     Also  Ri=P  and  at  =e,  hence 


,,    3PI     n       n  x 

max.  M=—. Pe=P(— —  e}. 

4  \4 


(23F) 


From  Eqs.  (23E)  and  (23r)  it  is  found  that  when  e>0.4494Z,  then  two  loads  give 
a  greater  max.  M  than  do  three  loads,  all  loads  being  equal. 

Case  IV.  For  four  equal  loads  on  the  span  may.  M  occurs  under  the  second  load 
and  Eq.  (23c)  gives 

max.M=P(l+—-2e]  ........  (23o) 


Here  again  it  is  found  that  when  e>0.2679Z,  then  three  loads  give  a  greater  max- 
imum than  four  loads. 

(b)  Critical  loading  for  maximum  moments.  Direct  loading.  In  this  case  the 
moment  influence  line  for  any  point  n  is  a  triangle  and  the  discussion  is,  therefore, 
applicable  to  any  beam  or  truss  for  which  the  moment  influence  line  is  a  triangle.  See 
Fig.  23B. 


:i 


n-~lD..O  01      n    h    n 


FIG.  23B. 

Let  R  =  2P  =the  total  load,  or  resultant  of  loads  on  the  span. 

RI  =the  resultant  of  all  loads  between  A  and  n  for  a  certain  position  of  a  train 

of  loads. 
RZ=  the  resultant  of  all  loads  between  n  and  B  for  the  same  position  of  the 

train  of  loads. 

Calling  rji  and  rj2  the  influence  ordinates  for  the  load  points  of  RI  and  R2  respec- 
tively, then  the  moment  for  the  point  n  may  be  written  as 

2—RiXi  tan  ai  +R2x2  tan  «2.    .     .     .     .     .     (23n) 


Suppose  now  that  the  train  is  moved  to  the  right  by  a  small  distance  dx,  producing 
the  change  +dx  in  Xi  and  the  change  —  dx  in  x2,  then  the  change  in  Mn  is  dMn  and  the 
differential  coefficient  thus  obtained  is  equated  to  zero  for  maximum  and  minumum, 
thus: 

dMn 


dx 


=Ri  tan  ai  —R%  tan  a2  =0. 


(23j) 


60  KINETIC  THEORY  OF  ENGINEERING  STRUCTURES         CHAP.  IV 

But  since  tan  «i  =^-.  and  tan  a2  =— ,  Eq.  (23j)  gives,  by  substitution, 
a  i  a-2 

^1=??.  (23K) 

ai      a2 ' 
By  observing  that  al  +a2=Z  and  ^i  +R2=R,  then  by  composition,  Eq.  (23K)  gives: 


_ 

a\  -f  02      a\      I 


Either  Eq.  (23K)  or  Eq.  (23L)  may  serve  as  a  criterion  for  the  critical  position  of 
the  train  load  on  the  span.  The  load  Pn,  falling  at  n,  is  divided  equally  between  1^ 
and  R-Z. 

In  the  first  instance  the  criterion  would  be  satisfied  when  the  average  unit  loads  on 
both  sides  of  the  point  n  are  equal,  a  condition  which  is  absolutely  fulfilled  by  a  uniformly 
distributed  load. 

According  to  Eq.  (23L)  the  criterion  for  max.  M  would  be  that  the  average  unit  load 
to  the  left  of  the  point  n  must  equal  the  average  unit  load  on  the  whole  span. 

Also,  since  the  maximum  ordinate  y  is  under  the  load  point  n,  no  position  of  loads 
can  give  a  real  maximum  unless  one  of  the  loads  is  directly  over  n.  Hence  it  is  readily 
understood  that  there  are  usually  two  or  more  maxima  for  any  point  n,  depending  on 
which  load  is  placed  over  this  point.  Therefore,  so  long  as  the  criterion  is  fulfilled,  all 
positions  giving  such  a  max.  Mn  must  be  used  to  determine  the  real  maximum. 

In  general  for  any  moment  influence  line,  Fig.  23s,  the  moment  will  be  maximum 
for  a  point  n  when  a  load  P  is  applied  at  n  and  when  the  angle  ^  of  the  influence  line 
exceeds  180°.  The  moment  at  n  will  be  minimum  only  when  there  is  a  load  at  some 
point  B  for  which  0'<180°,  that  is  where  the  angle  between  two  successive  elements 
of  the  influence  line  is  convex  upward.  Hence  all  angles  in  a  moment  influence  line 
represent  critical  load  points  and  correspond  to  maxima  or  minima  accordingly  as  these 
angles  are  convex  downward  or  upward,  respectively. 

(c)  Critical  loading  for  maximum  moments.  Indirect  loading.  For  any  point  w, 
other  than  a  panel  point,  the  moment  influence  line  is  a  quadrilateral,  hence  all  cases 
for  which  this  is  true  are  included  here. 

Let  .RI  —the  resultant  of  all  loads  acting  between  A  and  c. 
R%  =the  resultant  of  all  loads  acting  in  the  panel  ce. 
Rs  =the  resultant  of  all  loads  acting  between  e  and  B. 

Other  notation,  as  shown  in  Fig.  23c,  which  represnts  a  beam  or  girder  with  four 
panels,  each  of  length  d. 

From  Eq.  (23H)  the  general  expression  for  an  influence  line  of  any  number  of  sides 
may  be  written  as 

M  =  2Rx  tan  a,    ...........     (23M) 

and  the  differential  coefficient,  equated  to  zero  for  maximum  and  minimum,  becomes 

--  =  2R  tan  a  =0  ......     ...     .     .     (23u) 


ART.  23 


INFLUENCE   LINES  AND   AREAS 


61 


The  various  values  of  R  tan  a  are  now  found  from  the  special  case  in  hand,  using 
the  notation  and  illustration  in  Fig.  23c.  Three  such  terms  are  here  required  and  these 
are  evaluated  as  follows: 

jj  V  Tjc     a\  —d\        fje     a2  —d2 

tana-i=--;      tanas-—;       also  ~~^~' 

a\  a2  TJ          a\  T}          a2 


from  which  tana2  = 

With  #1,  /22  and  #3  and  these  values  of  on,  «2  and  «3,  and  observing  that  for  dx 
to  the  right  the  value  R3  tan  «3  must  be  negative,  then  Eq.  (23N)  becomes 


_ 
dx 


_ 
d 


(23o) 


FIG.  23c. 
which  may  be  reduced  to  give  the  criterion 


(23p) 


Observing  that  R=Ri+R2+Rs  as  before;    also  that  d=dl  +d2  and  that  l=( 
then  by  composition  of  Eq.  (23?)  and  substitution  of  these  values  the  second  form  of 

criterion  is  obtained  as 

d\ 

_!l=:? (23Q) 

a\  I 

Eq    (23r)  expresses  the  criterion  for  max.  Mn  when  the  average  unit  loads  on  both 
sides  of  the  point  n  are  equal,  provided  the  resultant  R2  is  distributed  between  R,  and  R*  in 

the  proportion  di :  d2. 

Eq    (23q)  stipulates  that  the  average  unit  load  over  the  distance  01  must  be  equal   to 
the  average  unit  load  over  the  whole  span,   provided  the  portion  R^/d  is   added    to   the 

resultant  R\.  .  . 

The  question  of  finding  which  particular  load  must  be  placed  at  either  c  < 
fulfill  either  of  the  above  criteria  may  be  solved,  by  trial,  as  for  the  case  of  direct  loading. 


62  KINETIC   THEORY  OF  ENGINEERING   STRUCTURES         CHAP.  IV 


ART.  24.     POSITIONS   OF   A   MOVING   TRAIN   FOR   MAXIMUM   AND   MINIMUM 

SHEARS 

(a)  Critical  loading  for  maximum  and  minimum  shear.     Direct  loading.     The  shear 
influence  line  is  then  as  shown  in  Fig.  ISA  and  the  max.  +Qn  is  obviously  produced  by 
a  full  loading  of  the  entire  span  between  B  and  the  load  divide  n.     The  min.—Qnis 
produced  by  a  full  loading  over  the  distance  An,  which  corresponds   to  the    negative 
influence  area.     This  is  absolutely  correct  for  uniformly  distributed  loads. 

For  a  concentrated  load  system,  the  train  must  cover  the  portion  of  span  between 
n  and  B,  with  the  first  load  just  to  the  right  of  n  so  that  its  influence  ordinate  is  nn"  without 
any  negative  effect  from  the  ordinate  nn'  .  However,  if  the  first  load  is  small  compared 
with  the  second  load,  it  may  necessitate  placing  the  second  wheel  at  n  to  attain  the 
real  max.  +Qn  =A  —Pi. 

Hence,  it  is  clear  that  either  PI  or  P2  must  be  just  to  the  right  of  n  to  obtain  max.  +Q, 
and  whichever  gives  the  greater  value  is  then  the  real  maximum. 

(b)  Critical  loading  for  maximum  and  minimum  shear.     Indirect  loading.      The 
influence  line  for  shear,  Fig.  19A,  or  the  stress  influence  line  for  a  web  member,  Fig.  21  A, 
are  both  included  here.     Hence  the  criterion  for  position  of  loads  for  a  max.  Q  will  also 
serve  for  maximum  stress  in  a  web  member. 

For  uniformly  distributed  loads  the  load  divide,  as  found  in  Fig.  20A,  will  always 
establish  the  point  at  which  the  head  end  of  the  load  must  be  located  for  max.  +Q  or 
min.  —Q. 

When  dealing  with  concentrated  load  systems  the  position  for  maximum  and 
minimum  shear  is  most  advantageously  found  by  graphics,  as  illustrated  in  Fig.  22e, 
where  the  magnitude  of  these  shears  is  at  once  determined,  together  with  the  critical 
position  of  loads. 

When  analytic  methods  are  employed,  the  following  criterion  may  serve  a  useful 
purpose. 

Let  Fig.  24A  represent  a  truss  with  bottom  chord  loaded  and  i  is  the  load  divide 
for  the  diagonal  D.  The  train  of  loads  covers  the  span  from  B  to  i,  making  RI  the  resultant 
of  the  loads  in  the  panel  CF  and  R2  the  resultant  of  the  loads  between  B  and  F.  The 
line  A'C'F'B'  is  the  stress  influence  line  for  the  member  £>. 

Then  the  stress  in  the  member  D  is 


(24A) 


By  shifting  the  train  dx  to  the  right,  both  Xi  and  x  are  diminished  by  this  amount 
dx.     Hence  the  stress  now  becomes 


ART.  24  INFLUENCE   LINES  AND   AREAS  63 

Subtracting  Eq.   (24s)   from  Eq.   (24A)   the  differential  change   in  SD  is  obtained 

dS  =SD -SD'  =rt \  -^,—  +^===.  | .  (24c) 


thus 


From  Eq.  (24 c)  the  value  dS/dx  is  found  and  equated  to  zero  for  maximum  value 

of  SD.     This  gives 

RI       R2  Ri+R2    Ri    R 

T,=l^^       or  by  composition       __=_=_,       .      .     .     (24D) 

where  R  is  the  resultant  of  all  loads  between  B  and  i.  This  furnishes  the  criterion, 
provided  no  additional  loads  have  entered  the  panel  CF  and  no  other  loads  have  come 
on  the  span  in  making  the  shift. 


FIG.  24A. 


It  is  evident  from  the  influence  line,  Fig.  24A,  that  any  loads  to  the  left  of  i  would 
produce  a  negative  influence  on  SD,  so  that  the  minimum  value  of  SD  may  be  found 
by  loading  the  left  portion  Ai  of  the  span,  and  the  criterion  then  becomes 

RI     R 


This  criterion  expresses  exactly  the  same  condition  as  previously  found  for  moments 
in  Eq.  (23L),  but  the  span  now  becomes  the  length  a\  or  a2,  as  the  case  may  be,  cover- 
ing only  that  portion  of  the  stress  influence  line  D,  which  has  the  same  sign. 

As  previously  found,  by  the  graphic  method,  it  is  always  necessary  to  place  a 
load  at  the  panel  point  F  for  max.  D  and  one  at  C  for  min.  D,  allowing  as  many  loads 
inside  the  panel  CF  as  may  be  required  to  satisfy  the  criterion. 


64  KINETIC  THEORY  OF  ENGINEERING   STRUCTURES          CHAP.  IV 

The  above  discussion  of  the  various  criteria  for  position  of  moving  loads  to  produce 
maximum  and  minimum  stresses  in  any  member  of  a  statically  determinate  structure 
is  ample  demonstration  of  the  high  value  and  practical  use  of  influence  lines. 

It  was  not  deemed  advisable  to  extend  this  discussion  to  cover  the  various  types 
of  special  and  composite  structures,  as  the  influence  lines  themselves,  when  used  for  the 
determination  of  stresses,  will  give  complete  and  comprehensive  answers  in  all  cases 
without  special  analytic  treatment.  However,  when  such  is  desirable,  the  methods 
previously  employed  will  indicate  the  procedure  to  be  followed  for  any  particular 
influence  line. 

The  subject  of  influence  lines  for  statically  indeterminate  structures  and  for  deflec- 
tions will  be  treated  after  presenting  the  general  subject  of  distortions  and  deflection 
polygons  of  framed  structures. 


CHAPTER  V 

SPECIAL   APPLICATIONS    OF    INFLUENCE    LINES   TO   STATICALLY  DETERMINATE 

STRUCTURES 

ART.  25.     DOUBLE    INTERSECTION   TRUSSES 

Usually  trusses  of  this  type  are  analyzed  as  two  separate  systems,  dividing  the 
loads  between  them  in  such  manner  as  may  seem  most  probable. 

The  stresses  found  for  the  separate  systems  are  then  combined  for  the  double  system 
wherever  the  same  member  forms  part  of  each  system,  provided  simultaneous  cases  of 
loading  were  used. 

The  method  of  influence  lines  has  advantages  especially  in  the  distribution  of 
simultaneous  load  effects,  and  the  combined  stress  in  a  member  belonging  to  both  systems 
is  at  once  found  for  the  same  position  of  loads. 

Double  intersection  trusses  should  always  be  designed  with  an  even  number  of 
panels,  thus  retaining  symmetry  of  both  systems  with  respect  to  the  center  of  the  span. 
Otherwise  the  loading  carried  by  one  system  in  the  left  half  span  must  go  to  the  other 
system  in  the  right  half  span,  which  is  not  desirable. 

Fig.  25A  represents  a  double  intersection  truss  for  which  stress  influence  lines  will 
now  be  drawn. 

Bottom  chord  member  DE.  A  section  tt,  passed  through  the  panel  DE,  cuts  the 
diagonals  FE  and  GK,  hence  the  chord  DE  forms  part  of  two  panels  for  which  there 
are  two  centers  of  moments,  F  and  G.  Therefore,  each  system  gives  rise  to  a  stress 
influence  line  for  the  member  DE,  shown  in  Fig.  25A,  by  the  two  dotted  triangles  A'C'B' 
and  A'D'B'.  The  top  chord  members  are  regarded  straight  between  alternate  pin  points 
for  each  system  in  question. 

Loads  acting  at  panel  points  are  supposed  to  be  carried  by  the  system  to  which 
each  such  point  belongs  except  at  the  panel  0  where  the  load  is  divided  equally  between 
the  two  systems. 

Hence  a  load  at  D  goes  entirely  to  one  system  and  its  influence  is  represented  by  the 
ordinate  DD' ',  of  the  influence  line  A'D'B';  while  a  load  at  E  is  carried  by  the  other  sys- 
tem and  its  influence  ordinate  becomes  EE'  in  the  influence  line  A'C'B'.  This  same 
reasoning  applies  to  all  panel  points. 

Since  influence  lines  between .  successive  panel  points  must  be  straight  lines,  the 
actual  influence  line  for  DE  is  represented  by  the  shaded  area  A'O'C'D'E'K'  to  B' . 
The  point  0'  must  lie  midway  between  the  two  ordinates  for  0  because  at  this  point 
the  load  is  carried  equally  by  both  systems.  This  is  not  the  case  at  M,  where  the  entire 
load  goes  to  one  system. 

65 


60 


KINETIC   THEORY  OF   ENGINEERING   STRUCTURES 


CHAP.  V 


Top  chord  member  LF.  The  influence  line  for  DE  is  like  the  influence  line  for  LF, 
except  that  the  end  ordinates  are  slightly  different.  Here  the  lever  arms  are  measured 
normally  to  the  members  LF  and  FG. 

Diagonal  member  FE.  The  method  of  reasoning  just  applied  to  chord  members 
answers  in  every  sense  for  the  web  members. 

The  influence  line  for  FE  is  constructed  as  indicated  in  the  figure.  The  stresses 
in  FE,  due  first  to  a  unit  load  at  A  and  then  at  B,  are  found  for  the  system  to  which  FE 
belongs. 


FIG.  25A. 


Since  loads  at  the  panel  points  of  the  other  system  cannot  cause  stress  in  FE,  the 
influence  line  ordinates  for  these  points  must  be  zero.  This  means  that  the  other  influence 
line  is  the  base  line  A'B'.  Observing  again  that  the  required  influence  line  must  be 
straight  between  successive  panel  points  the  final  line  becomes  A'O'C'D'E'K'  to  B'. 
The  point  0' ,  as  before,  is  midway  between  the  ordinates  under  0. 

The  influence  line  indicates  clearly  that  when  loads  are  spaced  2d  apart,  then  the 
entire  web  stress  goes  into  one  system,  making  the  design  verj'-  uneconomical.  This 
would  easily  happen  for  50  ft.  locomotives  and  25  ft.  panels. 


ART.  26  INFLUENCE  LINES  FOR  STATICALLY  DETERMINATE  STRUCTURES    07 


ART.  26.     CANTILEVER  BRIDGES 

Any  statically  determinate  structure  extending  over  more  than  two  supports  (piers 
or  abutments),  when  continuous  over  r—2  of  these  supports,  may  be  regarded  as  a 
cantilever.  If  r  is  the  total  number  of  supports,  then  such  a  structure  must  be  provided 
with  r—2  intermediate  hinges  such  that  the  bending  moments  at  these  hinged  points 
are  zero  for  any  system  of  loads. 

The  reaction  and  typical  stress  influence  lines  for  such  a  cantilever  will  now  be 
drawn. 

Reaction  influence  lines  for  all  simple  structures  are  independent  of  the  panels  and 
may,  therefore,  be  illustrated  on  a  simple  cantilever  beam,  Fig.  26A. 


FIG.  26A. 

The  illustration  represents  an  anchor  arm  A  B  with  a  downward  reaction  or  anchorage 
at  A  and  an  upward  pier  support  at  B.  The  span  EF  is  a  simple  beam  on  two  supports 
E  and  F,  with  an  overhanging  cantilever  arm  at_each_  end.  _A  similar  arm  projects  out 
from  the  anchor  arm,  making  three  cantilevers,  BC,  DE  and  FG,  in  the  whole  system. 

The  two  spans  CI)  and  GH  are  known  as  intermediate  spans  and  are  supported  on 
hinged  ends  which  permit  of  horizontal  displacements.  Hence,  there  is  only  one  roller 
bearing  at  E,  and  the  anchorage  at  A  also  admits  of  slight  horizontal  motion. 

The  whole  system  is,  therefore,  determinate  and  can  have  no  temperature  stresses, 
neither  will  slight  displacements  of  the  supports  cause  any  internal  stresses. 

The  reaction  influence  lines  thus  involve  no  principles  other  than  those  already 
elucidated  for  simple  beams  on  two  supports. 

The  central  span  WF  is  a  suitable  beginning,  as  it  rests  on  two  supports  R3  and  #4 
with  a  roller  bearing  at  E.  Hence  the  influence  lines  for  these  two  reactions  are  drawn 
precisely  as  for  a  simple  beam  between  the  points  E  and  F.  But  the  two  arms  at  the 
ends  of  this  span  must  also  be  considered. 

Regarding  first  the  influence  line  for  R3,  it  is  seen  that  a  load  at  F  has  zero  effect, 


68 


KINETIC  THEORY   OF  ENGINEERING    STRUCTURES 


CHAP.  V 


and  as  the  load  proceeds  to  the  left  its  effect  on  R%  is  uniformly  increasing  and  this 
continues  to  be  the  case  until  the  load  reaches  the  point  D,  where  the  effect  becomes 
a  maximum. 

The  same  is  true  for  a  load  between  F  and  G,  hence  the  ordinate  unity  at  E' 
determines  the  RS  influence  line  from  D^_io  G' .  A  load  at  C  or  at  H  produces  zero 
effect  on  Rs  and  loads  outside  the  span  CH  have  no  influence,  hence  the  line  C'D'G'H' 
is  the  influence  line  for  Rs.  Further  proof  will  scarcely  be  needed,  though  the  several 
moment  equations  for  the  separate  spans  CD  and  GH  will  readily  supply  this. 

Similarly  the  line  C"D"G"H"  represents  the  influence  line  for  R$,  based  on  the 
unit  ordinate  at  F". 

The  influence  line  for  Rs  is  the  same  as  for  a  simple  beam  GH. 

Precisely  the  same  reasoning  applies  to  the  reactions  Ri&nd  R-2,  the  influence  lines 
for  which  are  shown  to  the  left  in  Fig.  26A.  It  should  be  noted  with  regard  to  RI  that 
the  negative  influence  area  being  greater  than  the  positive,  provision  must  be  made 
for  a  downward  reaction.  When  the  upward  reaction  RI,  due  to  live  load  from  A  to 
B,  is  greater  than  the  downward  reaction  R'i,  due  to  dead  load  from  A  to  D,  then  the 
fastening  at  A  must  be  designed  to  take  stress  in  both  up  and  down  directions. 

Shear  influence  lines.  These  are  readily  deduced  from  the  reaction  influence  lines, 
as  may  be  seen  by  comparing  Fig.  26s  with  Fig.  26A.  Four  typical  panels  are  assumed. 
One  each  for  the  central  span  EF,  the  cantilever  arms  BC  and  FG,  and  the  anchor 
arm  A  B,  respectively. 


FIG.  26s. 


For  the  panel  ef  the  shear  influence  line  within  the  span  is  precisely  as  for  any 
simple  beam  on  two  supports,  while  outside  the  span  these  lines  are  continued  to  the 
intersections  D'  and  G'  with  the  verticals  through  the  hinges  D  and  G,  respectively. 
The  shear  influence  line  for  the  panel  ef  is  thus  found  as  C'D'e'f'G'H' . 

For  the  panel  ab  the  same  construction  is  applied  as  for  a  span  on  two  supports 
A  and  B  followed  by  a  cantilever  to  the  right  of  B  exactly  as  for  the  previous  case. 
The  shear  influence  line  for  the  panel  ab  is  the  polygon  A'a'b'B'C"D" '. 

When  the  panel  is  located  in  one  of  the  cantilever  arms  as  for  gh,  then  the  circum- 
stances are  slightly  different,  because  the  shear  in  any  cantilever  arm  is  always  equal 
to  the  sum  of  the  applied  forces  between  the  section  and  the  outer  end  of  the  arm. 


ART.  26   INFLUENCE  LINES  FOR  STATICALLY  DETERMINATE  STRUCTURES  69 


Accordingly  the  shear  must  be  composed  of  the  end  reaction  from  the  intermediate 

Hence  the  influence  line 
It  should  be  noted  that 


truss  GH  and  the  constant  effect  of  loads  between  h  and  G. 


for  the  panel  gh  is  easily  found  to  be  the  polygon  g'h'G"H' . 

loads  to  the  left  of  g,  even  on  the  next  span  EF,  can  have  no  effect  on  the  shear  in  the 

panel  gh. 

The  shear  influence  line  for  the  panel  cd  is  similarly  found  to  be  the  polygon  c'd'C"'D" '. 

Any  panel  located  within  either  of  the  intermediate  spans  CD  and  GH,  is  treated 
precisely  as  for  a  span  on  two  supports,  since  no  load  outside  these  spans  can  produce 
any  effect  on  these  trusses. 

Moment  influence  lines.  These  offer  very  little  difficulty  and  follow  directly  from 
the  methods  previously  given. 

In  Fig.  26c  the  anchor  arm  was  omitted,  as  the  portion  of  the  structure  from  C 
to  H  affords  all  the  illustration  required. 

Here,  again,  the  central  span  EF  is  treated  exactly  like  a  simple  beam  and  the 
negative  effects  of  the  adjoining  cantilever  arms  are  found  by  prolonging  the  lines  so 


FIG.  26c. 


obtained  to  the  intersections  D'  and  G'  with  the  verticals  through  the  hinged  points. 
The  moment  influence  line  for  the  point  n,  in  the  panel  EF,  thus  becomes  C'D'E'e'f'F'G'H' , 
and  it  is  readily  seen  that  a  load  on  any  portion  of  the  structure  CH  affects  the  moment 
at  n. 

For  the  section  t,  in  the  panel  gh,  the  moment  is  the  reaction  at  G  into  the  lever 
arm  a3,  hence  for  a  unit  load,  the  ordinate  at  G  must  be  l-a3,  and  observing  that  the 
influence  line  must  be  straight  over  the  panel  gh,  the  required  moment  influence  line 
for  the  section  t  becomes  F"g'h'G"H". 

Stress  influence  lines.  The  general  method  for  stress  influence  lines,  illustrated  in 
Fig.  21  A,  serves  every  purpose  in  connection  with  cantilevers. 

In  every  determinate  structure  the  stress  in  any  member  may  be  expressed  as 
S=M/r  and  from  this  it  follows  that  the  stress  influence  line  for  any  truss  member 
is  similar  to  the  moment  influence  line  drawn  for  the  center  of  moments  pertaining  to  the 
member.  The  difference  between  the  stress  and  moment  influence  lines  is  only  the 
constant  divisor  r  applied  to  the  ordinates  of  the  latter  to  obtain  the  ordinates  for  the 
stress  influence  line. 

For  convenience,  the  central  span  CE,  Fig.  26o,  will  be  treated  first,  because  this 


70 


KINETIC  THEORY  OF  ENGINEERING   STRUCTURES 


CHAP.  V 


portion  of  the  structure  is  exactly  like  a  simple  truss  on  two  supports,  and  all  the  rules 
previously  given  apply  here. 

Regarding  the  web  members  it  should  be  observed  that  the  center  of  moments  may 
fall  inside  or  outside  the  limits  of  the  span.  In  the  former  case  there  will  be  no  load 
divide  within  the  span.  Both  cases  are  illustrated  in  Fig.  26D,  where  the  stress  influence 
lines  for  members  U,  L,  D,  and  D'  are  presented. 

The  designations  regarding  the  stresses  in  these  members  for  unit  reactions  at  C 
and  E,  will  be  as  previously  adopted  in  describing  Fig.  2lA,  viz.,  Suc=ihe  stress  in  the 
member  U  for  a  reaction  unity  acting  at  C. 


FIG.  26o. 

Since  these  stresses  are  always  necessary  in  constructing  influence  lines,  it  is  best 
to  find  them  for  all  members  of  the  structure  before  proceeding  to  draw  any  influence 
.lines.  These  stresses  are  very  easily  found  either  by  slide  rule,  using  Ritter's  moment 
method,  or  by  drawing  a  Maxwell  diagram. 

Stress  influence  line  for  U,  central  span.  This  being  a  top  chord  member,  and  in 
compression,  its  influence  line  is  negative  between  C  and  E,  which  is  also  recognized  by 
the  stresses  —S^,  and  —  Sue  found  for  this  member  for  the  reactions  unity,  acting  first 
at  C  and  then  at  E. 

These  stresses  are  laid  off  from  A'B',  Fig.  26D,  upward  and  on  the  verticals  through 


ART.  20  INFLUENCE  LINES  FOR  STATICALLY  DETERMINATE  STRUCTURES    71 

C  and  E,  respectively.  The  stress  influence  line  is  then  easily  drawn,  observing  that 
the  intersection  n'  must  fall  under  the  center  of  moments  n  for  the  member  U.  Hence 
this  is  a  check  and  may  also  be  used  as  a  shorter  method  of  finding  the  U  line,  when 
only  —  Sllc  is  known.  Between  the  panel  points  m  and  o  the  influence  line  must  be  straight, 
hence  m'o'  is  drawn. 

To  complete  the  influence  line  over  the  side  spans,  the  U  line,  just  found,  is  continued 
to  intersections  B"  and  F"  with  the  verticals  through  the  hinged  points  B  and  F  and  the 
final  U  line  is  then  found  to  be  A'B"m'o'F"G' .  The  portions  below  the  base  line  A'G' 
are  positive  and  the  upper  portion  is  negative.  The  positions  of  loads  to  produce  max.  U 
and  min.  U  are  thus  clearly  indicated  and  no  other  criterion  is  required. 

Stress  influence  line  L,  central  span.  This  is  constructed  in  precisely  the  same  manner 
as  the  U  line,  but  L  being  in  tension,  as  found  for  C  =  l,  the  stress  +Sic  is  laid  off  down- 
ward under  C.  The  point  m" .  vertically  under  the  center  of  moments  m  for  the  member 
L,  is  thus  found  arid  may  serve  to  complete  the  L  line  without  using  the  stress  Sie. 
The  L  line  is  then  obtained  as  A'B'm"F'G'  and  is  positive  over  the  span  CE  and  negative 
over  the  two  cantilever  spans  AC  and  EG. 

Stress  influence  line  D,  central  span.  The  center  of  moments  for  this  member  is  at 
i,  which  is  within  the  supports,  and  hence  this  member  has  no  load  divide.  The  stress 
influence  line  is  constructed  exactly  as  was  done  for  the  U  line,  by  laying  off  the  positive 
stresses  Sdc  and  Sde  to  obtain  the  two  limiting  rays,  with  intersection  i'  under  the  point  i. 

However,  since  the  rate  of  increase  in  the  stress  D  must  be  uniform  for  a  load  advanc- 
ing from  E  to  o,  the  ray  E'r7i'  must  be  continued  to  oly  finally  drawing  the  line  m^i  to 
complete  the  polygon. 

Over  the  two  side  spans  the  D  line  is  drawn  as  for  the  two  previous  cases,  giving 
the  complete  line  A'"Bf"Cf"miolE'"F"'G"'. 

Stress  influence  line  D',  central  span.  The  center  of  moments  for  D'  is  outside  the 
span  at  k.  hence  the  influence  line  must  show  a  load  divide.  The  stresses  +  S'dc  and  —  S'de 
serve  to  construct  the  stress  influence  line  for  D'  which  is  in  all  other  respects  like  the 
D  line.  The  limiting  rays  intersect  in  k'  and  the  load  divide  is  at  q. 

Stress  influence  line  U,  cantilever  arm,  Fig.  26B.  The  effect  of  a  moving  load,  coming 
on  the  span  from  left  to  right,  begins  when  the  load  is  just  to  the  right  of  A  and  increases 
uniformly  until  the  load  reaches  B.  From  B  toward  m  the  effect  is  uniformly  decreas- 
ing and  becomes  zero  when  the  load  is  over  n,  which  is  the  center  of  momentsjor  U. 
The  stress  in  U  for  a  load  P  =  l  at  B  is-Sub,  and  laying  this  off  from  the_base  A'o',  the 
negative  U  line  is  easily  drawn.  For  the  effect  due  to  loads  in  the  panel  mo  draw  m'o'  to 
complete  the  influence  line.  Loads  to  the  right  of  o  or  to  the  left  of  A  have  no  influ- 
ence on  U. 

Stress  influence  line  L,  cantilever  arm.  The  center  of  moments  for  L  is  at  o  and 
the  stress  in  L  for  a  load  P  =  l  at  B  is+Slb.  Hence  the  influence  line  becomes  A"B"o" 

and  is  positive. 

Stress  influence  line  D',  cantilever  arm.  The  center  of  moments  for  this  diagonal 
is  at  i,  which  is  inside  the  cantilever  arm.  The  rule  for  load  divide  is  now  reversed 
because  loads  on  opposite  sides  of  i  produce  opposite  stresses  in  D',  and  a  load  at  i  cannot 
produce  any  stress.  Hence  i  is  the  load  divide  for  D'. 


72 


KINETIC   THEORY    OF   ENGINEERING  STRUCTURES 


CHAP.  V 


The  stress  in  D'  due  to  a  unit  load  at  F  is  —S'df,  hence  making  the  ordinate  under 
F  equal  to  this  stress  the  influence  line  s'i'F'G'  is  obtained. 

Since  loads  to  the  left  of  h  cannot  affect  D',  and  since  the  influence  line  is  a  straight 
line  over  a  panel,  the  influence  line  is  completed  by  drawing  h's'. 

Stress  influence  line  D,  cantilever  arm.  The  center  of  moments  now  falls  off  the 
arm  at  k  and  hence  all  loads  produce  the  same  kind  of  stress  in  D.  Lay  off  the  stress 
+Sdf  on  the  vertical  through  F  and  complete  the  influence  line  as  shown  in  Fig.  26E, 
obtaining  s"v'F"G"  for  the  D  line. 

The  intermediate  spans  A B  and  FG  are  simple  trusses  on  two  supports  and  require 
no  further  consideration. 


FIG.  26E. 


ART.  27.     THREE-HINGED   FRAMED   ARCHES 

In  taking  up  this  subject  it  is  necessary  to  show  briefly  the  general  relations  between 
the  reactions  and  external  loads. 

Let  Fig.  27A  represent  any  three-hinged  arch  ACB,  which  may  be  framed  or  solid. 
The  single  load  P  may  be  applied  at  any  point  m. 

For  all  positions  of  P  between  A  and  C,  the  reaction  R%  will  always  have  the 
direction  BC  and  the  two  reactions  RI  and  R%  must  be  components  of  P.  Similarly 
for  all  positions  of  the  load  P  between  C  and  B,  these  reactions  will  have  the  directions 
AC  and  eB,  respectively.  Hence  for  any  position  of  the  load  P,  the  reactions  at  A  and 
B  are  found  by  resolving  P  into  two  components  as  shown,  and  the  points  of  application 
of  all  possible  positions  of  P  will  lie  on  the  prolongation  of  either  AC  or  BC. 

If  the  reactions  R±  and  R2  be  resolved,  each  into  a  vertical  component  and  one 
along  AB,  the  two  last-named  components  H'  will  be  equal  and  opposite,  while  the  two 
vertical  components  A  and  B  will  be  precisely  the  same  as  for  a  simple  beam  on  two 
supports  A  and  B,  because  the  forces  acting  at  C  are  in  equilibrium  among  themselves. 

Therefore, 


A+B=P; 


Pa2 
A  —-, 


and       B=— =— . 


(27A) 


AKT.  27  INFLUENCE  LINES  FOR  STATICALLY  DETERMINATE  STRUCTURES   73 


The  horizontal  component  of  H'  is  known  as  the  horizontal  thrust  of  the  arch  and 
is  found  from 

H=H'cosa (27B) 

If  the  reactions  R\  and  R2  had  been  decomposed  into  components _4/,  B'  and  H, 
then  the  former  would  no  longer  be  the  reactions  for  a  simple  beam  AB.  However, 
when  a  =0  and  the  end  supports  are  on  a  horizontal  line,  then  the  vertical  end  reactions 
are  always  as  for  a  beam  on  two  supports. 


FIG.  27A. 

Taking  moments  about  the  hinge  C,  when  P  alone  is  acting,  then  for  condition  of 

equilibrium  at  C 

M=All-H'y=All-H'fcosa=Q, 

and  substituting  values  from  Eqs.  (2?A>  and  (27s)  this  gives 

Hi      or       H= — J-J-.  .     . 


(27c) 


When  P  =  l  is  applied  at  C  the  value  Hc  becomes  the  ordinate  vertically  under  C 
for  the  influence  line  for  horizontal  thrust. 
Hence, 

,-i,  •     •     •     (27D) 


-_^       and  when       Zi=Z2=o,       then 


making  the  H  influence  line   a  triangle  with  middle  ordinate  T?C  under  C,  as  given  by 

Eq.  (27D). 

The  influence  lines  for  the  vertical  end  reactions  are  found  exactly  as  for  a  simple 

beam  AB. 

The  stress  influence  line  for  any  member  of  a  three-hinged  framed  arch  will  now  be 

developed. 

When  the  structure  is  symmetric,  the  half  arch  only  requires  analysis,  but  for 
unsymmetric  arches  the  entire  structure  must  be  treated,  though  the  method  remains 
the  same  as  would  appear  from  the  previous  discussion. 


KINETIC  THEORY   OF  ENGINEERING  STRUCTURES 


CHAP.  V 


In  Fig.  2?B,  the  half  arch  is  shown  with  a  load  P  producing  reactions  R\  and  R2, 
and  since  the  latter  must  be  transmitted  to  the  point  C,  the  right  half  of  the  span  may 
be  considered  removed  and  its  effect  replaced  by  R2  acting  at  C.  The  left  half  of  the 
span  is  then  held  in  equilibrium  by  the  two  reactions  R:  and  R2)  the  latter  applied  at  C. 

The  reaction  R\  is  now  resolved  into  A  and  K  and  the  reaction  R2  is  similarly 
replaced  by  a  vertical  reaction  C  and  a  reaction  K  which  is  equal  arid  opposite  to  the 
first  K. 


FIG.  27s. 


If  the  forces  K  were  zero,  then  the  half  span  would  represent  a  simple  truss  on  two 
supports,  with  vertical  reactions  A  and  C  and  the  influence  line  for  any  member  could 
be  found  by  the  method  previously  given.  Thus,  the  influence  line  for  any  member- 
is  easily  drawn  when  the  stresses,  first  for  A=l,  and  then  for  5  =  1  are  known.  Such 
an  influence  line  is  drawn  for  the  diagonal  D,  and  is  shown  as  the  polygon  A'E'F'C' ',. 
Fig.  2?B,  where  the  stress  Da,  due  to  A  =  1  is  positive,  and  the  stress  Dc,  resulting  from 
C  =  l,  is  negative.  In  each  case  the  half  span  is  supposed  to  be  held  first  at  C  and  then 
at  A  in  order  to  find  these  stresses. 


ART.  27   INFLUENCE  LINES  FOR  STATICALLY  DETERMINATE  STRUCTURES  75 

The  influence  line  thus  found,  must  now  be  corrected  so  as  to  include  the  effect  of 
the  force  K. 

Since  K=Hsec{3,  therefore  K  is  proportional  to  H  and  the  stress  Dk,  produced 
by  the  force  K,  must  likewise  be  proportional  to  K  and  H.  Hence,  the  influence  line 
for  the  stress  Dk,  is  similar  to  a  horizontal  thrust  line  and  must  be  a  triangle  whose 
middle  ordinate  under  the  hinge  C  is  equal  to  the  stress  Dk,  produced  by  a  unit  load 
applied  to  the  whole  span  at  the  point  C. 

To  find  this  stress  Dk  for  a  unit  load  at  C,  substitute  P--=l  into  Eqs.  (27  A)  and 
(27c)  and  obtain 

A=i       and       #  =  l'i- 

J 

If  the  stress  in  D  is  found  first  for  A  =  1/2  and  then  for  //  =l/4f}  separately,  then 
the  combined  effect  will  be  the  stress  Dk=%Da+DH,  which  is  much  more  convenient 
than  to  find  Dk  directly  for  the  diagonal  force  K,  The  stress  DH  is  the  stress  in  D 
for  the  above  force  H  =  l-l/4f  applied  horizontally  at  A  and  supposing  the  half  arch 
rigidly  held  at  C. 

The  stress  Da  was  previously  found,  and  hence  the  required  stress  D^  is  readily 
obtained  from 


(27B) 

In  the  present  case  D^  was  found  to  be  negative  and  the  influence  line  for  the 
member  D  due  to  the  effect  K  is  drawn  as  the  triangle  A"C"B"  '  . 

The  final  D  line  is  now  obtained  by  combining  the  two  influence  lines  just  found- 
This  is  done  in  the  last  figure  and  gives  the  polygon  A'"E"F"C'"B"r.  The  stress  +Da 
is  laid  off  down  from  A'"  and  the  stresses  —  Dc  and  —  D^  being  negative,  are  laid  off 
upward  from  the  base  line  A'"B'"  and  being  of  the  same  sign  they  are  added.  The 
polygon  is  then  drawn  as  indicated  in  the  figure  and  as  a  check,  the  point  n"  must  fall 
vertically  under  the  'center  of  moments  n  for  the  member  D.  Still  another  check  is 
found  from  the  circumstance  that  a  load  at  i  produces  zero  stress  in  the  member  D  and 
hence  the  influence  line  must  intersect  the  base  line  vertically  under  i  in  the  point  i'. 
This  last  mentioned  condition  is  always  true  even  though  the  point  i  falls  below  C. 

The  point  i  is  the  intersection  between  BC  and  An,  where  n  is  the  center  of  moments 
for  D.  This  offers  a  very  valuable  condition  and  obviates  the  necessity  of  finding  the 
stress  Dc,  whenever  the  center  of  moments  n  can  be  located  within  limits  of  the  drawing. 

It  is  thus  seen  that  when  n  and  i  are  conveniently  located,  which  is  always  the  case 
for  chord  members,  then  the  influence  line  can  be  determined  when  Da  alone  is  given. 
However,  as  a  check,  it  is  always  desirable  to  know  D^,  so  that  when  the  stresses  in 
all  the  members  are  to  be  determined  by  influence  lines,  it  is  advisable  to  find  Sa  and 
Sh  for  each  member  before  proceeding  to  draw  the  lines.  This  can  be  done  by  drawing 
two  Maxwell  diagrams,  one  for  A  =  1  and  the  other  for  H=l/4f.  A  slide-rule  computa- 
tion will  also  answer  if  the  lever  arms  are  carefully  determined. 

In  Fig.  27s  the  loads  were  supposedly  applied  to  the  top  chord.  Had  they  been 
applied  to  the  bottom  chord  then  the  panel  line  E"F"  would  have  become  G"H". 


76 


KINETIC  THEORY  OF  ENGINEERING  STRUCTURES 


CHAP.  V 


Typical  stress  influence  lines  will  now  be  given  for  the  three-hinged,  framed  arch, 
Fig.  27c,  treating  the  top  chord  as  the  loaded  chord. 

The  example  will  be  taken  up  in  such  manner  as  to  indicate  a  complete  solution 
for  all  stresses  in  the  several  members  by  the  method  of  influence  lines. 


~?B 


B' 


I     UU    INFLUENCE  LINE. 

"^luiii]|p|i||]^ 


90  LIN&THV 


The  first  step  will  be  to  determine  the  stresses  Sa  and  Sk  for  all  the  members  and 
this  will  be  done  analytically  by  Ritter's  method  of  moments,  being  very  careful  about 
the  sign  of  each  stress. 

The  same  stresses  could  also  be  found  in  a  very  convenient  way  by  drawing  two 
Maxwell  diagrams,  one  for  A~l,  supposing  the  half  truss  ~AC  fixed  at  C;  and  the  other 


ART.  27  INFLUENCE  LINES  FOR  STATICALLY  DETERMINATE  STRUCTURES   77 


for  a  load  P  =  l  acting  vertically  downward  at  C  and  producing  the  horizontal  thrust 
//  =  l-Z/4/= 0.833,  which  is  again  applied  at  A,  supposing  the  half  truss  fixed  at  C. 

The  following  table  contains  the  stresses  Sa  and  Sa  thus  obtained  and  the  stresses 
S=Sa+S    are  then  easil    found. 


MEMBER. 

STRESSES. 

MEMBER. 

STRESSES. 

-Sa 

SH 

8t 

Sa 

SH 

Sh 

U,A 

-1.00 

+  1.02 

+  0.52 

U& 

-0.45 

+  0.52 

+  0.30 

f7,L, 

+  0.93 

-0.57 

-0.11 

iW 

-1.37 

+  1.08 

+  0.40 

u& 

-1.29 

+  0.79 

+  0.15 

U3Ut 

-3.11 

+  1.97 

+  0.42 

U2L2 

+  1.32 

-0.85 

-0.19 

ALt 

0.00 

-1.44 

-1.44 

U3L2 

-1.65 

+  0.77 

-0.05 

L,L. 

+  0.85 

-1.87 

-1.45 

U3L3 

+  2.19 

-1.15 

-0.06 

L2L3 

+  2.08 

-2.38 

-1.34 

UtL3 

-2.45 

+  0.96 

-0.26 

L3C 

+  4.40 

-3.35 

-1.15 

utc 

+  3.05 

-1.19 

+  0.33 

+  for  tension  and  —  for  compression. 

The  values  in  this  table  will  suffice  for  the  construction  of  all  the  stress  influence 
lines,  but  only  those  for  (72t/3,  L2L3,  U2L2  and  £/4L3  will  be  drawn,  as  shown  in  Fig. 
27c. 

Stress  influence  line  U2U3.  The  stress  Sa=— 1.37  is  laid  off,  to  a  convenient  scale, 
upward  from  A'  and  the  stress  £^=+0.40  is  laid  off  downward  from  A'B',  vertically 
under  C.  The  lines  EC'  and  C'B'  are  then  drawn. 

The  center  of  moments  for  U2U3  is  at  L2,  hence  find  Z/2  vertically  under  L2  and 
Finally  since  the  influence  line  must  be  straight  over  the  loaded  panel 

As  a  check  observe 


draw  A'L2'. 
UjU~3,  draw  U2'US 


to  complete  the  influence  line  A'U2UzC'B'. 


that  the  load  divide  i/  falls  vertically  under  i',  which  latter  is  the  intersection  of  AL2 
with  BC  produced.  Still  another  check  may  be  had  by  resolving  a  unit  load  at  U2 
into  components  parallel  to  U2US  and  U2L2  as  was  done  to  the  right  of  the  truss  diagram. 
The  stress  thus  found  for  U2U3  must  equal  the  ordinate  U2'F.  Lastly,  the  ordinate 
— Sc  is  the  stress  in  U2U3  for  a  load  unity  at  C  when  the  half  truss  AC  is  fixed  at  A 
and  the  half  span  CB  is  removed. 

Stress  influence  line  L2L3.  The  method  is  precisely  as  for  the  previous  case,  being 
careful  to  observe  the  signs  of  the  stresses  Sa  and  Sk  in  laying  them  off  from  the  base 
line  A7/?'.  The  point  U'3,  vertically  under  the  center  of  moments  for  this  member, 
again  serves  to  complete  the  stress  influence  line,  with  all  the  checks  just  mentioned 
for  the  upper  chord  member. 

Stress  influence  lines  U2L2  and  U4L3.  The  same  method  again  applies  and  the  two 
lines  illustrate  the  effects  of  the  signs  of  the  stresses  Sa  and  Sk.  The  center  of  moments 
n,  for  the  member  U2L2  now  falls  outside  the  span,  but  the  same  relations  hold  good 
as  before.  Similarly  for  the  center  HI  of  the  member  U4L3.  In  this  last  case  the  load 
divide  t'3  falls  below  the  middle  hinge  C  and  is  no  longer  a  real  load  divide,  as  the  influence 
line  for  TT^Ls  indicates,  but  all  other  relations  continue  to  exist  as  before. 


KINETIC  THEORY  OF   ENGINEERING   STRUCTURES  CHAP.  V 


ART.  28.     THREE-HINGED,    SOLID-WEB    AND   MASONRY   ARCHES 

Fig.  28A  illustrates  a  three-hinged,  solid-web  arch,  which  may  be  built  of  masonry 
or  steel,  according  to  circumstances.  If  of  masonry,  then  the  section  will  be  rectangular, 
and  if  of  steel,  then  the  section  will  resemble  that  of  a  plate  girder. 

Whenever  the  depth  of  the  section  is  small  compared  with  the  radius  of  the  arch 
center  line,  then  the  effects  of  curvature  are  very  minute  and  the  stresses  in  the  arch 
rib  may  always  be  found  by  the  well  known  beam  formula  of  Navier.  This  condition 
always  prevails  in  bridges  and  all  larger  arches,  but  when  dealing  with  small  arched 
culverts  and  structures  of  that  class,  then  the  effects  of  curvature  may  require  special 
consideration. 

For  any  girder  section  then,  the  stress  on  the  extreme  fiber  is  derived  from 


N 

wherein  M  is  the  bending  moment,  A"  the  normal  axial  thrust,  F  the  cross-section,  / 
the  moment  of  inertia  of  the  section  and  ?/  the  distance  from  the  center  of  gravity  of 
the  section  to  the  extreme  fiber.  Mk  is  the  total  bending  moment  of  the  external 
forces  about  one  of  the  outer  kernel  points  e  or  i  (Fig.  28A)  and  W  is  the  moment  of 
resistance  of  the  section. 

Calling  r  the  radius  of  gyration  of  the  section,  then 

7     Fr2  r2  I 

W=-=  —  =Fk       where       k=—^-~  ......     (2SB) 

y     y  y       Fy 

The  quantity  k  is  the  distance  from  the  gravity  axis  to  a  kernel  point  and  the  negative 
sign  indicates  that  it  is  measured  in  the  opposite  direction  from  y  for  all  sections. 

Hence,  the  influence  lines  for  the  kernel  moments  Me  and  Mi  will  serve  to  find  the 
stresses  fe  and  /;  for  the  extreme  fibers  of  extrados  and  intrados  of  any  particular  section 
tl,  by  using  the  multiplier  /y.  =  l/W  =  l/Fk.  For  unsymmetric  sections  the  two  kernel 
distances  are  not  equal  so  that  {ie  =  l/Fke  for  extrados  and  fi.i  =  \/Fki  for  intrados. 

For  each  section  the  stresses  on  the  extreme  fibers  thus  become 

''-7S,     and    fi=~K     .....   '    •    '    (28c) 

where  Me=Kace  and  Mi=Kaci,  and  Kac  is  the  end  reaction  at  A  for  a  load  unity  at  C 
found  by  resolving  the  vertical  reaction  A  along  AC  and  AB.  See  also  Art.  49  on  two- 
hinged  arches. 

The  moment  influence  lines  Me  and  M;  are  now  found  by  the  method  previously 
outlined  for  framed  arches. 

Each  of  these  moments  becomes  zero  when  the  unit  moving  load  passes  through 
a  load  divide  d.  Hence  the  load  divides  are  easily  located  for  each  case. 

Also  since  moments  and  not  stresses  are  now  considered,  the  end  ordinate  for  the 
influence  line  at  A  is  equal  to  xe  or  Xi  as  the  case  may  be. 


ART.  28   INFLUENCE  LINES  FOR  STATICALLY  DETERMINATE  STRUCTURES  79 


Hence  to  draw  the  influence  line  Mi,  lay  off  Xi  at  A  and  project  the  point  d  down 
to  d',  draw  Gd'C'  and  after  projecting  down  the  center  of  moments  i  to  find  i',  finally 
complete  the  influence  line  by  drawing  A'i'  and  C'B'.  Now  find  as  a  check,  that 
~C'~D  =  x'i. 


\ 


B' 


FIG.  28A. 


The  Aft-  influence  line  becomes  the/;  influence  line  with  a  factor  m=* 

The  influence  line  Me  is  found  in  precisely  the  same  manner. 

In  both  cases  the  ordinate  C'C"  is  equal  to  Mt  or  Me  as  given  by  Eq.  (28c)  for  a 

unit  load  at   C.     The   Me  influence  line  becomes   the  /«   influence   line  with   a  factor 


80  KINETIC  THEORY  OF  ENGINEERING  STRUCTURES  CHAP,  v 

The  shear  influence  line  QOT  is  found  by  a  similar  process.  Here  the  center  of 
moments,  for  a  web  member  at  m,  would  be  practically  at  an  infinite  distance  because 
the  chords  are  parallel  or  nearly  so.  Hence  the  line  AdAj.t  will  intersect  the  chord 
C/I  in  the  load  divide  d,  which  in  this  case  is  only  imaginary,  but  serves  the  purpose 
as  before. 

The  equation  for  shear  along  the  section  it  is  easily  found  to  be 


and  this  gives,  for  a  unit  reaction  at  A,  when  the  half  arch  is  fixed  at  C,  the  end  ordinate 
A'G  =  l-cos  <f>.  Also  since  the  shear  is  constant  between  A  and  m,  the  limiting  lines 
A'm'  and  Gd'  must  be  parallel. 

Hence  the  shear  influence  line  is  easily  constructed  by  drawing  GC'd'  ,  and  then 
completing  the  polygon  by  drawing  A'm'  ||  Gd'  and  joining  C'  with  B'  '  . 

Since  the  loading  is  directly  applied  the  line  Wira',  vertically  under  m,  completes 
the  Q7>,  line. 

Finally  the  ordinate  C'C"  =H[  —  —  -  -  )  wherein  7/  =  l--^rfor  a  unit  load  at  C. 

\      cos  a      )  fl 

When  the  loads  are  indirectly  applied,  then  the  condition  that  the  influence  line 
must  be  straight  between  load  or  panel  points  must  be  carefully  observed. 

For  masonry  arches  the  section  is  always  rectangular  and  7  =/>D3/12,  hence  k  =  —r'2/y  = 
^F^Z),  where  D  is  the  thickness  of  the  arch  ring.  The  points  e  and  i  thus  become  the 
middle  third  points  of  the  arch  section  and  Eqs.  (28c)  become 

,     GMe  , 

f*~~TF     and     ^= 

from  which  the  stresses  in  a  masonry  arch  section  are  readily  found  in  terms  of  the 
moments  about  the  kernel  points  of  the  section.  The  stress  influence  lines  thus  become 
moment  influence  lines  with  a  factor  6/D2  for  each  section. 

ART.  29.     SKEW  PLATE   GIRDER   FOR   CURVED   DOUBLE   TRACK 

In  general,  influence  lines  for  skew  bridges  are  drawn  in  all  respects  like  those  for 
any  simple  truss  or  beam  with  the  exception  that  at  one  end  of  the  span  the  load  always 
leaves  the  span  before  it  is  entirely  outside  of  the  limits  of  the  longer  truss.  This  merely 
calls  for  a  slight  correction  in  the  end  panel  of  the  influence  line  as  ordinarily  drawn. 

At  the  opposite  end  of  such  truss  some  load  comes  on  the  end  floor  beam  before 
the  moving  load  is  on  the  truss,  but  since  the  end  floor  beam  transmits  its  load  directly 
to  the  bridge  support  and  not  to  the  truss,  this  last  named  circumstance  does  not  affect 
the  influence  line.  These  points  will  be  brought  out  in  connection  with  the  following 
problem. 

It  may  seem  inappropriate  to  apply  influence  lines  to  plate  girders,  but  the  practical 
designer  who  has  dealt  with  problems  of  skew  plate  girders  on  curved  double  track 
will  readily  appreciate  the  advantages  of  the  method  there  presented. 


KT.  29  INFLUENCE  LINES  FOR  STATICALLY  DETERMINATE  STRUCTURES   81 


MOMENT  INFLUENCE  LINE  "  ~  -- 
POINT  7,  GIRDER  NO.I.TRACK  NO.2. 


FIG.  29x. 


82  KINETIC   THEORY  OF  ENGINEERING  STRUCTURES  CHAP.) 

This  method  was  first  suggested  and  used  by  Mr.  T.  B.  Moenniche,  C.  E.,  in  designing 
a  skew  double-track  girder  of  125  ft.  span  for  the  Virginian  Ry.  Co.,  in  1906.  Thi 
author  is  indebted  to  Mr.  Moenniche  for  the  solution  which  follows. 

The  principal  advantage  in  the  present  application  of  influence  lines  consists  ii 
reducing  the  solution  of  this  rather  complicated  problem  to  a  method  instead  of  a  time 
consuming  process  of  experimentation. 

The  principle  of  the  method  may  be  outlined  as  follows:  The  floor-beam  reaction! 
are  first  determined  for  a  unit  load  situated  on  track  Xo.  1,  and  then  for  a  unit  load  01 
track  No.  2,  which  requires  a  careful  computation  of  the  coordinates  or  intersectioi 
points  of  each  track  on  each  floor  beam.  These  floor  beam  reactions  are  then  used  a! 
influence  or  reduction  factors  for  the  ordinates  to  any  particular  influence  line,  becaus< 
the  factors  represent  the  proportion  of  a  total  load  which  can  reach  either  girder  a 
any  panel  point.  This  will  give  two  influence  lines  for  each  panel  point  of  a  girder 
one  separately  for  each  track. 

In  Fig.  29A,  the  upper  diagram  represents  the  structure  in  plan  and  the  two  sue 
ceeding  figures  show  the  floor  beam  reactions  graphically.  The  last  two  diagram: 
are  the  moment  influence  lines  for  point  7,  girder  No.  1,  for  tracks  No.  1  and  2,  respectively 

The  two  reaction  diagrams  were  drawn  merely  as  a  convenient  way  of  showing 
the  figures  and  in  practice  they  would  serve  the  same  purpose,  though  they  might  b( 
dispensed  with  entirely.  The  reactions  appearing  as  ordinates  are  plotted  to  sonn 
convenient  scale  making  the  sum  of  two  reactions  for  each  floor  beam  equal  to  unity 
except  at  the  ends  where  the  floor  beams  do  not  connect  with  both  girders. 

The  moment  influence  lines  for  point  7  are  drawn  in  the  usual  manner  by  laying 
off  a2  vertically  down  from  B  and  drawing  A'T  and  7'B' '.  Since  the  load  on  track  \o 
1,  when  coming  on  from  the  right,  begins  to  affect  girder  No.  1  as  soon  as  it  passes  the 
abutment  at  E,  therefore,  the  distance  #10  must  be  treated  as  a  panel  arid  the  influence 
line  is  drawn  straight  from  10  to  B'. 

The  ordinates  d  of  the  influence  line  are  now  successively  reduced  by  multiplying 
them  with  the  corresponding  ordinates  or  factors  m  of  the  reaction  diagram  to  obtaii 
the  real  ordinates  y  of  the  required  moment  influence  line.  Thus  ^6=<Vn6  and  similar!} 
for  the  influence  line  point  7,  girder  1,  track  No.  2,  the  ordinate  Jjg=08w8. 

With  the  use  of  these  two  influence  lines,  represented  by  shaded  areas,  the  actual 
moment  for  point  7  may  be  found  for  any  case  of  loading  for  each  track  and  the  sun: 
of  the  two  gives  the  combined  max.  M7. 

Shear  influence  lines  may  be  drawn  in  a  similar  manner,  though  they  would  scarcely 
serve  a  useful  purpose  except  when  dealing  with  a  truss.  In  the  above  example  onl} 
the  reaction  influence  lines  would  be  useful  to  determine  the  end  shears. 


AKT.  :-o  INFLUENCE  LINES  FOR  STATICALLY  DETERMINATE  STRUCTURES    S3 


ART.  30.     TRUSSES   WITH   SUBDIVIDED   PANELS 

From  a  point  of  economy,  large  trusses  must  have  long  panels,  but  with  the  increase 
in  panel  length  the  stringers  of  the  floor  system  soon  attain  such  proportions  as  to 
impose  a  practical  limitation  on  the  panel  length. 

To  overcome  this  difficulty,  modern  practice  has  developed  a  new  type  of  truss 
especially  adapted  to  long  spans,  wherein  the  stringers  are  supported  by  floor  beams 
placed  at  the  half-panel  points.  These  intermediate  floor  beams  are  carried  at  the  ends 
by  a  secondary  set  of  truss  members,  which  latter  serve  to  carry  the  loads,  thus  locally 
applied,  to  the  main  truss  system.  This  has  given  rise  to  the  truss  with  subdivided 
panels. 

In  the  analysis  of  these  truss  types,  it  is  not  always  a  simple  matter  to  determine 
the  criterion  for  maximum  stresses  in  the  members,  and  it  is  believed  that  the  following 
treatment  by  influence  lines  will  serve  to  clear  up  the  doubts  usually  encountered. 

Fig.  30A  shows  a  truss  of  200  ft.  span  with  four  different  forms  of  subdivided 
panels.  These  are  all  combined  in  the  one  structure,  though  in  practice  only  one  form 
should  be  used  in  the  same  structure  excepting  in  the  first  panels  ALiL2,  where  the 
several  compression  members  shown  offer  distinct  advantages  in  stiffening  the  ends 
of  the  truss.  Otherwise  the  panel  L2L3,  Case  I,  is  the  best  and  most  economical  type 
for  adoption,  as  will  appear  later. 

In  general,  all  primary  members  in  the  structure  will  receive  stresses  which  are 
at  least  equal  to  those  which  would  exist  if  the  secondary  members  were  all  removed 
and  in  addition  to  these  primary  stresses  nearly  all  of  the  members,  excepting  the 
verticals,  will  receive  increased  stresses  due  to  local  loads  from  the  secondary  members. 
This  local  loading  is  transmitted  to  the  truss  through  the  vertical  MZK  which  never 
acts  as  a  primary  member  and  gets  no  stress  other  than  that  due  to  a  full  panel  load, 
besides  the  dead  weight  of  bridge  floor  and  bottom  chords. 

It  is  the  load,  locally  transferred  to  the  points  M ,  which  enters  as  a  disturbing 
element,  and  the  four  cases  here  treated  will  illustrate  about  all  the  practical  points. 

To  make  the  four  cases  directly  comparable,  the  same  panel  L2L3  was  retained, 
but  the  diagonals  were  successively  rearranged.  The  stress  influence  line  for  the 
primary  stress  in  U2LS  remains  the  same  in  each  case,  hence  it  is  drawn  only  once 
and  is  then  modified  according  to  local  conditions  in  the  panel  L2L3  for  each  case. 

The  Diagonal  Member  U  ,L  ,. 

Case  I.  The  stress  influence  line  for  U2L3,  when  acting  only  as  a  principal  truss 
member,  is  drawn  by  laying  off  the  end  ordinate  A' A"  =  +0.568  which  is  the  stress 
U2L3  for  a  reaction  unity  at  A.  The  ordinate  B'B"  (not  shown)  is  —2.790  and  is 
the  stress  in  the  same  member  for  a  unit  reaction  at  B.  By  drawing  A" ' B'  and  A'B" 
the  points  U'2  and  L'3  are  found  and  the  line  t/2'L3'  completes  the  influence  line 
AfU2Ls'B'  with  load  divide  at  i.  

This  gives  the  maximum  stress  for  M3L3  but  not  for  U2MZl  because  a  load  at  K 
would  produce  additional  stress  in  the  latter  member  without  affecting  the  former. 

Hence  to  find  the  influence  line  for  U2M3,  it  is  necessary  to  determine  the  stress 


84 


KINETIC  THEORY  OF  ENGINEERING  STRUCTURES 


i     STRESS  INFLUENCE  LINE  M3L3. 


FIG.  30A. 


ART.  :so  INFLUENCE  LINES  FOR  STATICALLY  DETERMINATE  STRUCTURES  85 


in  this  member  due  to  a  local  load  unity  at  M3  which  gives  the  total  ordinate 
under  M3. 

A  load  at  Mz  will  be  carried  entirely  by  U2M3  and  M^U~3,  hence  the  stress  due  to 
a  unit  load  is  readily  found  by  resolving  the  unit  load  into  components  parallel  to  these 
two  members  as  indicated  by  the  triangle  of  forces  in  the  figure  for  Case  I,  drawn  to  half 
the  scale  of  the  influence  line  ordinates.  The  stress  in  U2M3  is  thus  found  to  be 
+0.5,  which  is  laid  off  from  m  to  complete  the  influence  line  A'U2'M3L3B'  for  the 
member  U2M3.  It  is  seen  that  the  point  M'3  does  not  fall  on  the  line  B'A"  as  usually 
assumed  by  authors  on  this  subject.  The  error  from  this  assumption  is  more  apparent 
in  Case  II,  and  is  not  always  on  the  safe  side. 

Case  II.  The  influence  line  for  the  primary  stress  in  U2L3  is  as  before,  and  again 
gives  maximum  for  the  lower  portion  M3L3.  However,  a  unit  load  at  M3  is  now  carried 
to  the  members  U2M3  and  L2M3,  and  these  stresses  are  again  found  by  a  triangle  of 
forces  drawn  in  the  figure  for  Case  II. 

The  stress  thus  produced  in  U2M3  is  now  +0.62,  from  which  the  new  influence 
line  for  U2M3  is  readily  obtained  as  A'U2'M3L3B'. 

Case  III.  Here  the  primary  stress  holds  good  for  the  upper  portion  U2M3  and 
the  stress  in  the  lower  portion  M3L3  is  diminished  by  the  compression  due  to  local 
loading  at  M3.  A  unit  load  at  M3  is  now  carried  by  M3L3  and  M3U3  and  the  triangle 
of  forces  again  gives  the  ordinate  iM3,  which  is  here  negative. 

Case  IV.  This  is  like  Case  III,  but  a  local  load  at  M3  is  now  carried  by  L2M3 
and  M3LS  equally.  Hence  the  stress  in  M3L3  due  to  a  unit  load  at  M3  is  found  as 
—0.62,  which  gives  the  ordinate  iM3. 

Members  L2M3  and  M3U3,  whenever  they  occur  alone,  can  act  only  as  secondary 
members  to  carry  the  portion  of  load  locally  applied  at  M3.  When  the  panel  is  subject 
to  counter  stress  as  in  panels  L4L5  and  L5Le,  then  the  local  stress  must  be  combined 
with  the  counter  stress  when  a  counter  U^MQ  is  employed.  However,  it  is  usually 
preferable  to  omit  the  counter  member  and  design  the  main  diagonal  to  take  both  the 
direct  and  counter  stresses. 

The  verticals  U2L2,  U3L3,  etc.,  may  be  treated  entirely  as  primary  members  assuming 
the  panels  to  be  L2L3}  L3L4,  etc. 

The  vertical  UjLi  is  merely  a  suspender  for  the  double  panel  AL2  and  its  maximum 
stress  will  occur  when  the  span  is  fully  loaded  from  A  to  L2. 

The  member  M6V,  and  others  of  similar  character  often  employed  in  large  bridges, 
are  entirely  secondary  and  perform  no  work  as  truss  members.  The  sole  purpose  of 
these  members  is  to  stiffen  certain  long  compression  members. 

Bottom  chord  member  L2L3.  The  influence  line  for  this  member  is  easily  drawn. 
The  center  of  moments  is  at  U2  and  the  end  ordinate  at  A  is  the  stress  in  the  member 
for  a  unit  reaction  at  A.  The  influence  line  thus  becomes  a  triangle  A'L2'B'. 

Case  I.  With  the  diagonals  all  as  tension  members  the  local  loads  at  the  middle 
points  cannot  affect  the  bottom  chord  stresses  and  the  influence  line  just  described 
gives  maximum  stress. 

Case  II.  Here  the  bottom  chord  stress  is  increased  by  the  local  load  transmitted 
from  M3  to  L2.  This  is  shown  in  Fig.  30A.  For  a  unit  load  at  M3  the  stress  in  the 


86  KINETIC  THEORY  OF  ENGINEERING  STRUCTURES  CHAP.  V 


bottom  chord  is  the  horizontal  component  in  L2M3  found  in  the  small  force  diagram 
for  Case  II.  This  stress,  equal  to  +0.36i_must  then  be  the  additional  plus  ordinate 
under  M3  and  the  stress  influence  line  for  L2L3  thus  becomes  A'L2K'L3B'.  as  shown. 

Case  III  is  exactly  like  Case  II,  and  the  horizontal  component  of  M3L3  due  to  a 
unit  load  at  M3  will  be  the  positive  ordinate  to  be  added  to  the  ordinary  influence  line 
under  K.  

Case  IV.  Here  the  two  members  L2M3  and  M3L3  will  transmit  the  unit  load 
from  M3  after  the  manner  of  a  three-hinged  arch.  The  local  stress  in  the  oottom 

L2L3  20 

chord  then  becomes  =====  = =  +0.358  for  a  unit  load  at  M3. 

4M3K     4X13.95 

Hence  the  positive  ordinate  under  K  must  be  increased  by  0.358  to  determine  the 
proper  influence  ordinate  under  K. 

Top  chord  member  U2U3  has  four  cases  as  just  described  for  the  bottom  chord. 

Case  I.  Here  the  local  load  at  M3  produces  a  compression  in  the  top  chord  like 
a  suspension  s}rstem.  The  amount  of  this  compression  for  a  unit  load  at  M3  is  readily 
found  from  the  Eq.  (27o)  given  for  three-hinged  arches  and  is  H'  =l\L2/fd  cos  a,  where 
li  +I2=d=pa,nel  length,  see  the  truss  diagram  Fig.  30A. 

This  will  give  the  negative  stress  ordinate  to  be  added  to  the  negative  stress 
ordinate  under  M3  for  the  final  influence  line. 

Cases  II  and  III  are  similar  and  here  the  local  effect  in  the  top  chord  is  the  com- 
ponent in  U2M3  or  M3U3  parallel  to  U2U3. 

Case  IV  does  not  influence  the  top  chord  for  local  load  at  M3.  It  is  similar  to 
Case  I  for  bottom  chord. 

It  is  thus  seen  that  the  best  criterion  for  trusses  of  this  type  is  obtained  from 
influence  lines  and  the  ease  and  clearness  with  which  any  case  is  solvable  speaks  greatly 
in  favor  of  this  method  of  analysis,  where  critical  positions  of  loads  and  stresses  are  all 
given  from  the  one  solution. 

In  conclusion  it  may  be  remarked  that  the  type  of  panel  illustrated  in  Case  I  is  in 
all  respects  preferable  to  the  others.  The  type  Case  II  deserves  second  choice,  and 
the  others,  especially  Case  IV,  should  be  avoided  whenever  possible. 

The  panel  L\L2  is  in  a  certain  sense  indeterminate  and  should  never  be  used  except 
in  end  panels  as  here  illustrated,  where  the  several  posts  tend  to  steady  the  shock  of 
trains  coming  on  the  structure. 


CHAPTER    VI 
DISTORTION    OF  A   STATICALLY   DETERMINATE    FRAME    BY  GRAPHICS 

ART.  31.     INTRODUCTORY 

The  displacement  between  any  two  points  or  pairs  of  lines  of  any  determinate 
frame  may  be  found  analytically  as  shown  in  Arts.  6  and  9.  However,  many  problems, 
to  be  considered  later,  require  a  complete  solution  for  every  pin  point  of  a  structure, 
and  then  the  analytic  method  would  become  quite  laborious. 

The  graphic  solution  here  given  includes  the  two  following  problems:  (1)  To  find 
the  distortion,  or  change  in  form,  of  a  frame  resulting  from  changes  in  the  lengths  of  its 
members,  first  published  in  1877  by  the  French  engineer  Williot;  and  (2)  to  find  the 
effect  on  this  distortion  produced  by  a  yielding  of  the  supports,  or  reaction  displacements. 
This  latter  contribution  came  from  Professor  Otto  Mohr  in  1887,  and  without  it  the  Williot 
diagram  had  little  real  value.  Hence  it  is  proper  to  call  the  complete  diagram  a  Williot- 
Mohr  displacement  diagram,  contrary  to  the  position  taken  by  Professor  Mueller- 
Breslau,  who  erroneously  credits  the  entire  subject  to  Williot. 

Having  given  the  changes  in  the  lengths  of  all  the  members  of  a  structure,  by  Eq. 
(4fi),  when  the  stresses  and  cross-sections  are  known,  let  it  be  required  to  find  the  hori- 
zontal and  vertical  displacements  of  all  the  pin  points. 

The  Williot  diagram,  to  be  presented  first,  offers  a  partial  solution  of  this  problem, 
which  is  completed  by  Mohr's  rotation  diagram. 

ART.  32.     DISTORTIONS   DUE   TO   CHANGES   IN   THE   LENGTHS   OF   THE 
MEMBERS  BY  A  WILLIOT  DIAGRAM 

Since  any  determinate  frame  is  formed  by  a  succession  of  triangles,  this  elementary 
frame  is  first  examined.  The  process  may  then  be  extended  to  include  any  number  of 
such  triangles  or  a  complete  frame. 

In  Fig.  32A  suppose  the  point  c  to  be  connected  with  the  points  a  and  6  by  members 
1  and  2,  and  that  the  lengths  of  these  members  also  undergo  changes  —  Al  and  +  J2, 
respectively,  while  the  points  a  and  b  move  to  new  positions  a'  and  &'.  It  is  now 
required  to  find  the  direction  and  amount  of  displacement  resulting  for  the  point  c. 

Both  members  are  first  moved  parallel  to  themselves  until  they  occupy  the  positions 
a'c2  and  b'ci}  respectively.  The  length  of  member  1  is  now  shortened  by  the  amount 
Jl  because  this  is  a  negative  change;  similarly  the  member  2  is  lengthened  by  the 
amount  J2,  giving  the  new  lengths  a'c4  and  b'c3,  respectively. 

87 


8S  KINETIC  THEORY  OF  ENGINEERING  STRUCTURES          CHAP.  VI 

With  a'  and  6'  as  centers  and  corrected  lengths  a'c4  and  b'c3  as  radii,  describe  arcs 
which  may  intersect  in  a  point  c'  representing  the  new  position  of  the  point  c.  But 
because  Al  and  Al  are  always  very  minute  compared  with  the  lengths  of  the  members 
to  which  they  belong,  the  arcs  c4c'  and  c3c'  are  replaced  by  their  tangents,  drawn  respec- 
tively perpendicular  to  the  original  directions  of  the  members.  The  diagram  shows 
these  displacements  several  hundred^  times  larger  than  they  actually  are  compared 
with  the  real  lengths  of  the  members  ac  and  be. 

It  is  clear  that  the  point  c'  might  have  been  determined  in  a  separate  figure, 
without  including  the  members  themselves,  by  dealing  exclusively  with  the  directions 
and  changes  in  the  lengths  of  these  members.  Thus,  in  Fig.  32B,  the  Jl's  are  plotted 
on  a  very  large  scale,  affording  greater  accuracy  in  the  results. 

If  the  point  0  is  regarded  as  a  fixed  point  or  pole,  and  if  from  this  pole  the  dis- 
placements of  the  points  a  and  6  are  drawn,  making  Oa'  =aa'  and  Ob'  =66',  both  in 
magnitude  and  direction;  also,  applying  at  a'  the  shortening  —  Jl  in  the  length  of  the 
member  1;  and  at  b'  the  lengthening  +J2  of  the  member  2;  then  the  perpendiculars 


FIG.  32B. 


erected  at  the  extremities  of  A\  and  J2  will  intersect  in  c' ,  which  is  the  new  position 
of  c.  The  displacement  of  c  is  then  represented  in  direction  and  amount  by  the  ray  Oc'. 

In  laying  off  the  values  of  Al,  in  Fig.  32s,  the  following  rule  must  be  observed  with 
regard  to  signs:  If  a'  is  regarded  as  fixed,  then  —Al  representing  the  shortening  of  ac, 
it  follows  that  c  moves  in  the  direction  from  c  toward  a,  and  hence  —  Jl  must  be  drawn 
from  a'  in  the  direction  of  c  to  a.  Likewise,  if  6'  is  regarded  fixed,  then  since  +J2 
represents  a  lengthening  of  be,  it  follows  that  c  moves  away  from  6  and  hence  +J2 
must  be  drawn  from  6'  in  the  direction  of  6  to  c. 

Any  succession  of  triangles,  as  in  Fig.  32c,  may  be  treated  in  the  same  manner. 
It  is  necessary,  however,  to  assume  that  one  of  the  members,  as  ab,  retains  its  direction, 
and  that  some  point  of  ab,  as  a,  remains  fixed.  The  changes  Al  in  the  several  members 
being  given,  the  construction  is  carried  out  as  follows : 

The  point  0,  Fig.  32c,  is  the  assumed  pole,  and  since  the  point  a  is  considered 
fixed  it  must  coincide  with  0.  Also,  since  the  member  ab  does  not  change  in  direction, 
then  —  Jl  is  drawn  parallel  to  ab  and,  being  negative,  it  must  be  applied  in  the  direction 
of  6  toward  a.  The  displacement  of  the  point  6  is  thus  given  by  Ob'. 


ART.  33        DISTORTION   OF  A  STATICALLY  DETERMINATE  FRAME  89 

The  point  c  recedes  from  a  by  an  amount  +J3,  and  approaches  b  by  an  amount 
—  J2.  If,  therefore,  J2  be  drawn  from  b'  in  the  direction  of  c  toward  b  and  parallel 
to  cb,  and  if  J3  be  drawn  from  0  or  a',  in  the  direction  of  a  toward  c  and  parallel  to 
ac,  then  the  intersection  c'  of  the  perpendiculars,  to  the  extremities  of  J2  and  J3,  will 
determine  the  new  position  of  c.  The  displacement  of  c  is  thus  given,  both  indirection 
and  amount,  by  a  ray  Oc'  not  drawn  in  the  figure. 

The  point  d  is  joined  to  a  and  c  by  the  members  ad  and  dc,  Fig.  32c.  Its  displace- 
ment Od'  is  now  found,  in  Fig.  32o,  by  drawing  +J4  parallel  to  ad  from  0  and  in  the 
direction  a  to  d;  also  by  drawing  —  J5  parallel  to  erf  from  c'  in  the  direction  d  to  c. 
The  intersection  d' ,  of  the  two  perpendiculars  to  the  extremities  of  J4  and  J5,  is  the 
new  position  of  d.  Finally  the  new  position  e'  of  the  point  e  is  similarly  found  by 
applying  the  same  method. 


I 
\ 

\ 
\ 
\ 
\ 


FIG.  32o. 


Fig.  32o,  whose  polar  rays  Ob',  Oc'  ,  Od'  ,  and  Oe'  give  the  displacements  of  the  several 
points  b,  c,  d  and  e,  both  in  direction  and  magnitude,  is  called  the  "Williot  displace- 
ment diagram,"  for  the  frame  abcde,  giving  credit  to  the  name  of  its  originator. 


ART.    33.     ROTATION    OF   A    RIGID   FRAME    ABOUT   A    FIXED   POINT— PROFESSOR 

MOHR'S  ROTATION  DIAGRAM 

In  the  Williot  diagram  it  was  assumed  that  the  member  ab  did  not  alter  its  position, 
but  merely  changed  its  length.  However,  this  is  not  always  the  case  and  more  frequently 
this  member  ab  is  also  subjected  to  a  rotation  about  some  instantaneous  center.  In 
the  latter  event,  the  elastic  displacements,  given  by  the  Williot  diagram,  will  require 
further  changes  to  make  them  comply  with  the  initial  change  in  the  position  of  the 
member  ab.  It  is  supposed,  for  the  present,  that  the  frame  has  undergone  its  elastic 
deformation  and  has  passed  into  a  rigid  state. 

The  solution  of  this  phase  of  the  problem  was  first  proposed  by  Professor  Mohr  and 
is  based  on  the  following  theorem  in  mechanics:  The  motion  of  a  rigid  body,  at  any 
given  instant,  may  be  defined  as  a  rotation  about  a  certain  point  called  the  instantaneous 
center  of  rotation,  and  the  direction  of  motion  of  any  point  of  this  body,  at  the  instant  in 


90 


KINETIC  THEORY  OF   ENGINEERING  STRUCTURES 


CHAP.  VI 


question,  will  be  perpendicular  to  the  line  joining  such  point  with  the  instantaneous  center 
of  rotation. 

Thus,  if  abcde,  Fig.  33A,  represents  a  rigid  frame  rotating  about  the  instantaneous 
center  P,  each  arrow,  perpendicular  to  its  respective  ray,  will  then  represent  the  direction 
of  motion  of  the  point  to  which  the  ray  is  drawn. 

Now,  if  from  any  pole  0,  Fig.  33B,  a  series  of  radial  rays  be  drawn  respectively 
parallel  to  the  arrows  representing  the  motions  of  the  several  points,  then  these  rays 
Oa",  Ob",  Oc",  Od",  and  Oe"  will  in  turn  represent,  both  in  magnitude  and  direction, 
the  several  displacements  of  the  several  points,  a  b,  c,  d  and  e.  The  figure  a"b"c"d"e" 
will  be  the  new  position  of  the  rigid  frame  abcde. 

For,  a"0-LaP;  b"0-LbP;  c"0-LcP;  etc.,  because  the  direction  of  motion  of  each 
such  point  of  the  rigid  frame  was  perpendicular  to  the  ray  joining  this  point  with  the 
instantaneous  center  of  rotation. 

Also.  a"0:b"0:c"0,  eic.=aP:bP:cP,  etc.,  because  the  displacements  of  the  several 
points  a,  b,  c,  etc.,  are  respectively  proportional  to  their  velocities,  and  these  in  turn 


FIG.  33A. 


FIG.  33n. 


are  proportional  to  the  distances  of  the  several  points  from  the  instantaneous  center 
of  rotation. 

From  these  conditions  it. follows  that: 

(a)  If  the  points  a",  b",  c" ,  etc.,  of  the  diagram  Fig.  33B,  are  joined  by  straight 
lines,  so  that  for  every  member  ab  of  the  frame,  there  will  be  a  corresponding  line  a"b" 
in  the  displacement  diagram,  then  these  latter  lines  will  produce  a  figure  a"b"c"d"e" 
which  will  be  similar  to  the  rigid  frame. 

(6)  Also,  the  line  joining  any  two  points  of  the  rigid  frame,  as  ab,  will  be  perpen- 
dicular to  the  corresponding  line  a"b"  of  the  displacement  diagram. 

Hence,  if  it  is  possible  to  determine  the  position  of  any  two  of  the  points  a",  b", 
c",  etc.,  of  the  displacement  diagram,  then  the  figure  a"b"c"d"e"  can  be  drawn  by- 
similarity  with  the  figure  abcde.  The  method  of  determining  two  such  points  a" ',  b" 
will  be  illustrated  in  the  examples. 

Therefore,  the  secondary  displacements,  due  to  rotation  of  the  rigid  frame,  are 
found  by  inserting  in  the  Williot  displacement  diagram  a  figure  a"b"c"d"e"  similar 
to  the  frame  abcde,  and  having  its  sides  respectively  perpendicular  to  the  sides  of  the 


ART.  M        DISTORTION  OF  A  STATICALLY  DETERMINATE  FRAME  91 

latter  frame.  The  original  displacements  Ob',  Oc',  etc.,  are  then  combined  with  the 
displacements  Ob",  Oc",  etc.,  due  to  rotation  to  produce  the  resultant  displacements 
V'J',  cr'7,  etc. 

The  complete  solution  for  the  displacement  of  the  pin  points  of  any  framed 
structure,  definitely  supported,  may  then  be  outlined  in  the  three  following  steps: 

(a)  Assume  any  member  and  a  point  on  the  axis  of  this  member   as    fixed    and 
construct  a  Williot  displacement  diagram. 

(b)  Then  assume  the  frame  as  rigid  and  subjected  to  a    rotation    such    as   will 
conform  to  the  actual  conditions  of  the  supports. 

(c)  The  displacements  of  the  several  pin  points  may  then  be   found,  in   magnitude 
and  direction,  by  scaling  the  distances  between  the  m"  and  m'  points.     The  direction 
of  any  displacement  will  always  be  from  m"  toward  m' '. 

The  complete  diagram,  combining  the  Williot  displacement  diagram  with  the  Mohr 
rotation  diagram,  will  hereafter  be  known  as  a  Williot-Mohr  diagram. 

The  analytic  solution  of  deflection  problems  is  conducted  with  the  aid  of  Professor 
Mohr's  work,  Eq.  (GA),  and  may  be  advantageously  employed  when  only  one  point  of 
the  structure  is  to  be  treated,  or  when  great  accuracy  is  required  in  the  solution. 

This  method  is  fully  discussed  in  Art.  6  and,  in  combination  with  Maxwell's  theorem 
given  in  Art.  9,  offers  the  most  accurate  analytic  solution  of  a  great  variety  of  problems 
dealing  with  displacements  of  points  and  lines. 

However,  all  problems  requiring  a  complete  solution  for  all  the  pin  points  of  a 
structure,  may  be  solved  by  a  Williot-Mohr  displacement  diagram  when  extreme 
accuracy  is  not  necessary. 

Attention  is  called  to  the  fact  that  the  value  for  E,  in  all  deflection  problems, 
should  be  chosen  low  rather  than  high,  because  there  is  always  a  slight  amount  of  lost 
motion  and  permanent  set,  which  follows  the  first  loading  of  a  new  structure  and  which 
is  not  truly  an  elastic  deformation.  This  can  be  compensated  for  in  the  calculations 
by  assuming  a  small  E. 

As  a  general  rule,  the  changes  Al,  in  the  lengths  of  the  members,  are  best  figured 
on  gross  sections  rather  than  on  net  sections.  Experience  indicates  that  this  gives 
results  nearer  the  truth.  Also  the  stresses  in  the  members  must  be  simultaneous  and 
not  maximum. 

ART.  34.     SPECIAL   APPLICATIONS   OF   WILLIOT-MOHR   DIAGRAMS 

(1)  Distortion  of  a  simple  truss.  Fig.  34A  has  been  solved  in  three  ways,  for  the 
purpose  of  illustrating  that  the  displacements  are  not  affected  by  the  choice  of  the 
member  and  point  which  are  regarded  as  fixed  in  position,  and  also  in  order  to  show 
how  to  select  the  most  convenient  of  the  possible  forms  of  solution. 

The  truss  is  supported  on  rollers  at  e  and  is  fixed  at  a.  The  extensions  (+)  and 
contractions  ( — )  of  the  members  are  assumed  to  be  as  marked  upon  them. 

The  solution  given  in  diagram  6  is  made  under  the  supposition  that  the  direction 
of  the  member  af  and  the  point  a  remain  fixed. 

According  to  the  method  outlined  in  Art.  32,  a  displacement  diagram  is  constructed 


92 


KINETIC  THEORY  OF   ENGINEERING   STRUCTURES          CHAP.  VI 


x    DOUBLE  *C/UE.  / 


y 


/(d)\ 


OF  DISPLACE MCNTS. 

i  ,  ,  ,  i  ,  .  .  ,  i ,,,,11 

o  i  2! 

FIG.  34A. 


ART.  34        DISTORTION   OF  A  STATICALLY  DETERMINATE  FRAME  «J3 

with  pole  at  0.  Since  a  is  now  supposed  to  remain  fixed,  the  point  a'  coincides  with 
0,  and  the  displacements  of  the  various  points  6,  c,  d,  e  and  /  are  obtained  by  scaling 
the  distances  Ob7,  Ocr ,  Od'  .  .  .Of. 

In  reality,  however,  the  direction  of  the  member  uf  does  not  remain  fixed,  for  the 
member  revolves  about  the  point  a.  Therefore,  the  displacements  just  found  must 
be  combined  with  those  displacements  06",  Oc",  Od",  etc.,  which  are  caused  by 
revolving  the  rigid  frame,  or  truss,  abge  about  the  point  a  until  the  point  e  shows  a 
resulting  displacement,  e'e" ',  parallel  to  the  direction  of  motion  of  the  roller  bed  at  e. 

In  other  words,  the  resultant  movement  of  the  point  e  will  be  horizontal  when 
the  roller  bed  is  horizontal.  If  the  bed  were  inclined,  as  em,  diagram  a,  then  the  point 
e  would  move  parallel  to  em. 

The  figure  a"b"c"d"e"f"g"h",  similar  to  the  figure  of  the  truss,  can  then  be  con- 
structed, making  the  sides  respectively  perpendicular  to  the  members  of  the  truss. 
This  figure  is  definite  and  can  be  drawn  when  the  points  a"  and  e"  are  found.  The 
point  a"  will  coincide  with  a'  or  0,  since  it  remains  fixed  and  a"e"  will  be  perpendicular 
to  ae.  Also  e'e"  will  be  parallel  to  the  roller  bed,  which  is  horizontal  in  the  present 
case,  but  may  be  in  any  direction,  as  e'm" ',  for  a  skew-back. 

The  actual  displacements  of  the  points  6,  c,  d,  etc.,  will  then  be  given  in  direction 
and  in  amount  by  the  distances  &"&',  c"c',  d"df,  etc.,  but  the  horizontal  and  vertical 
projections  of  these  displacements  are  more  generally  desired  and  may  be  scaled  from 
the  drawing. 

The  deflection  polygon  of  the  bottom  chord  is  found  graphically  by  projecting 
the  points  a',  a",  6',  b",  etc.,  on  to  the  verticals  through  the  panel  points  a,  6,  etc. 
The  points  a",  6",  c",  etc.,  will  be  projected  in  a'",  6"',  c'",  etc.,  and  will  form  a  straight 
line  a'"e'" .  The  points  6',  c' ,  d' ,  etc.,  will  fall  in  60,  c0,  d0,  etc.,  and  the  ordinates  606"', 
C(,c'" ,  etc.,  will  give  the  vertical  deflections  of  the  panel  points  6,  c,  etc. 

In  diagram  c  the  direction  cb  is  assumed  and  the  point  c  is  fixed.  All  that  has 
been  said  regarding  the  first  solution  applies  here.  It  will  be  seen  that  this  solution 
gives  exactly  the  same  displacements  as  previously  found,  while  it  occupies  only  about 
half  the  space  of  the  first  diagram  because  the  member  be  has  a  lesser  angular  motion 
than  the  member  af. 

The  third  solution,  diagram  d,  is  the  simplest  of  all,  and  its  diagram  covers  the 
least  area;  for  the  member  eg,  which  is  now  assumed  as  fixed, j-eally  has  no  angular 
motion,  but  simply  drops  vertically.  The  relative  displacement  b'g'  of  any  two  points, 
6  and  g,  may  be  scaled  off  directly.  Although  this  displacement  diagram  was  drawn 
on  the  assumption  that  the  line  eg  and  the  point  c  remained  fixed,  nevertheless  any 
two  points  may  be  directly  compared.  Hence,  any  point  may  be  chosen  as  the  origin 
of  coordinates  from  which  to  scale  the  horizontal  and  vertical  displacements  of  all  other 
points  relatively  to  this  origin.  Naturally  the  displacements  desired  would  be  those 
with  respect  to  the  point  a,  for  this  is  the  fixed  point,  giving  for  the  point  e  a  horizontal 
movement  from  a'  to  e'.  All  other  points  move  to  the  right  and  down  from  their 
original  positions  by  amounts  which  may  be  scaled  from  the  diagram,  taking  a'  for  the 
origin. 

Hence  it  may  be  concluded  that  if  a  truss  contains  a  member  which  will  move 


94 


KINETIC   THEORY  OF  ENGINEERING  STRUCTURES 


CHAP.  VI 


parallel  to  itself,  or  which  has  no  angular  motion,  then  this  member  is  the  best  one  to 
choose  as  the  fixed  member  in  constructing  the  displacement  diagram.  The  rotation 
displacements  are  thereby  eliminated. 

(2)  A  French  roof  truss.  Fig.  34s.  In  this  example  the  stresses  were  found 
from  a  Maxwell  diagram,  and  the  corresponding  changes  in  the  lengths  of  the  members 
were  obtained  by  means  of  the  formula  Jl=Sl/EF,  taking  E  at  25,600,000  pounds  per 
square  inch  for  wrought  iron.  Changes  of  length,  due  to  temperature,  are  neglected. 
The  values  of  Al  being  very  small,  ten  times  these  values  were  taken.  All  the  data 
necessary  in  the  solution  of  this  truss  are  given  in  the  following  table: 

TABLE   OF  THE   VALUES   OF  S,  F,  I,  AND  10JZ  IN  FIG.  34n 


Member. 

Stress, 
Lbs. 

S 

Area, 
Sq.  in. 
F 

Length, 
Inches, 
I 

10UO, 
Inches. 

Member. 

Stress, 
Lbs. 
S 

Area, 

Sq.  in. 
F 

Length, 
inches. 
I 

10(40, 
Inches. 

1 

-21,800 

6.82 

83.53 

-0.106 

15 

-20,970 

6.82 

83.53 

-0.102 

2 

-20.040 

6.82 

83.53 

-0.094 

16 

-19,210 

6.82 

83  .  53 

-0.091 

3 

-  18,300 

6.82 

83.53 

-0.087 

17 

-17,470 

6.82 

83.53 

-0.083 

4 

-16.540 

6.82 

83.53 

-0.079 

18 

-15,710 

6.82 

83.53 

-0.075 

5 

+  19,730 

2.02 

96.53 

+  0.370 

19 

+  15,330 

2.02 

96.53 

+  0.288 

6 

+  16,430 

2.02 

96.53 

+  0.307 

20 

+  13,570 

2.02 

96.53 

+  0.256 

7 

+   9,040 

1.86 

100  .  08 

+  0.189 

8 

+   9,260 

1.86 

96.53 

+  0.189 

22 

+  6,200 

1.86 

96.53 

+  0.126 

9 

+  12,560 

1.86 

96.53 

+  0.256 

23 

+   7,960 

1.86 

96  .  53 

+  0.161 

10 

-   3,300 

1.40 

48.07 

-0.043 

24 

-   1,680 

1.40 

48.07 

-0.024 

11 

+   3,300 

0.78 

96.53 

+  0.162 

25 

+   1,680 

0.78 

96.53 

+  0.087 

12 

-  6,600 

2.48 

96.53 

-0.102 

26 

-   3,520 

2.48 

96.53 

-0.055 

13 

+   3,300 

0.78 

96.53 

+  0.162 

27 

+    1,680 

0.78 

96.53 

+  0.087 

14 

-   3.300 

1.40 

48.07 

-0.043 

28^ 

-   1,680 

1.40 

48.07 

-0.024 

The  truss  is  composed  of  the  two  frames  aeh  and  sqh,  designated  as  I  and  II. 
These  are  connected  by  the  member  as  and  by  the  pin  at  h. 

The  displacement  diagram  of  frame  I  is  first  drawn,  assuming  as  fixed  the  direction 
of  any  member,  as  ab,  and  the  position  of  a  point,  as  a,  of  this  member.  The  point  a' 
then  coincides  with  the  pole  0,  and  the  displacement  Ob' ,  of  the  point  b  will  be  equal 
to  J12.  The  points  c' ,  d',  e*,  f,  g'  and  h'  are  then  found  as  directed  in  Art.  32. 

The  displacement  diagram  of  frame  II  is  next  drawn,  assuming  as  fixed  the  direction 
of  the  member  sk  and  the  position  of  the  point  s. 

Having  thus  determined  the  points  k' ,  I',  m',  q' ,  ri ',  p'  and  h'  in  diagram  II,  the 
relative  changes  in  the  positions  of  the  points  h,  s  and  q,  parallel  to  the  straight  lines 
joiningjthe  points  h  and  s,  h  and  q,  and  s  and  q,  may  be  found.  These  changes  are  called 
Ahs,  Ahq  and  Asq. 

Diagram  I  may  now  be  completed  by  inserting  the  values  Ahs,  Ahq  and  Asq  previously 
found  from  diagram  II.  The  point  s'  is  found  from  Al  and  Ahs,  and  the  point  q'  is 
found  from  Ahq  and  Asq.  Since  q  moves  on  a  horizontal  roller  bed  and  since  e  is  fixed, 
the  figure  c"d"b"g"h"f"a"s"q"  can  be  drawn  as  in  the  preceding  problem.  This  figure 
gives  the  displacements  of  all  the  points  of  frame  I,  and  those  of  the  points  s  and  q. 


ART.  34       DISTORTION  OF  A   STATICALLY  DETERMINATE  FRAME 

The  displacement  diagram  of  frame  II  may  now  be  completed  by  transferring  the 
displacements  of  the  points  h  and  q  from  diagram  I  into  diagram  II,  thus  determiniaf 


Lj_ 
loo 


«\       DIAGRAM  I. 

\        N 


DISPLACEMENTS. 

i 


DIAGRAM  H. 


too  zoo 

FIG.  34s. 


the  points  q"  and  h"  from  which  the  figure  q"l"s"ri'h"m"k"p"  can  be  drawn. 

As  a  check,  it  should  be  remembered  that  the  line  q"h"  in  diagram  II  must  be 


96 


KINETIC   THEORY   OF   ENGINEERING  STRUCTURES         CHAP,  vi 


perpendicular  to  the  line  hq,  and  that  the  displacement  s's"  of  the  point  s  must  be  the 
same,  both  in  direction  and  in  amount,  in  both  diagrams. 

Diagram  II  might  have  been  dispensed  with  in  the  present  case,  as  the  values  dhst 
Ahq  and  Asq  might  have  been  found  directly  by  summing  the  J/'s.  However,  the  use 
of  diagram  II  is  general,  and  it  becomes  necessary  when  the  points  h,  p,  k,  m  and  q, 
or  h  n  and  s,  or  /,  s  and  q,  are  not  in  straight  lines,  as  in  the  case  of  a  curved  top  chord. 

Measurements  from  diagram  I  show  that  the  point  g  undergoes  the  greatest 
displacement,  having  a  horizontal  movement  of  1.93/10=0.193  inch  to  the  right, 
a  vertical  downward  movement  of  2.33/10  =0.233  inch,  or  a  resultant  movement  of 
oV7  =3.02/10  =  0.302  inch.  The  horizontal  movement  of  the  point  q  is  =  q'q"  =2.61/10  = 


FIG.  34c. 


0.261  inch.  The  displacements  given  by  the  diagram  are  here  divided  by  10,  because, 
as  already  stated,  the  changes  Al  were  originally  taken  ten  times  too  large. 

(3)  A  three-hinged  arch.  Fig.  34c.  It  is  required  to  draw  the  displacement 
diagram. 

Independent  diagrams  for  each  of  the  elastic  frames,  I  and  II,  are  first  drawn  by 
assuming  as  fixed  the  direction  of,  and  a  point  on,  some  member  of  each  frame,  as  was 
done  in  the  previous  problem. 

The  frames  I  and  II  are  then  regarded  as  rigid,  and  each  frame  is  supposed  to 
revolve  in  such  a  way  as  to  satisfy  the  conditions  imposed  by  the  supports. 

The  displacement  diagrams  are  omitted  and  only  the  second  part  of  the  problem 
is  solved. 


ART.  34        DISTORTION  OF   A  STATICALLY  DETERMINATE  YRA  ME  97 

Supposing  that  the  points  a'  and  c'  in  Fig.  34c,  diagram  I,  and  the  points  b'  and 
c',  diagram  II,  have  been  found  from  the  corresponding  displacement  diagrams  (not 
shown) ,  let  it  be  required  to  complete  the  solution  by  adding  Mohr's  rotation  diagram. 

The  following  conditions  must  exist  between  the  figures  a"c"  and  b"c" ,  which  are 
to  be  similar  to  the  frames  I  and  II  respectively: 

a.  The  displacement  of  the  point  a  is  zero,  hence  a"  will  coincide  with  a'. 

b.  Similarly  b"  will  coincide  with  b'. 

c.  The  line  a"c",  in  diagram  I,  must  be  perpendicular  to  the  line  ac. 

d.  The  line  b"c" ,  in  diagram  II,  must  be  perpendicular  to  the  line  be. 

e.  Both  diagrams,  I  and  II,  must  give  the  same  displacement  c'c"  for  the  point  c. 

Hence  a"c"  is  drawn  through  a',  perpendicular  to  ac,  also  b"c"  through  b'  perpen- 
dicular to  be.  Now,  in  diagram  I,  c'n  is  drawn  parallel  to  ac,  intersecting  a"c"  in  n, 
and  the  projection  of  the  required  displacement  c'c",  parallel  to  ac,  is  thus  obtained. 
Likewise,  in  diagram  II,  c'm  is  drawn  parallel  to  cb,  intersecting  b"c"  in  m,  giving  c'm, 
the  projection  of  c'c"  parallel  to  cb. 

Diagram  I  may  now  be  completed  by  transferring  c'm  from  diagram  II  and  erecting 
a  perpendicular  to  c'm  at  m.  This  perpendicular  will  intersect  na"c"  in  c",  and  the 
figure  a"c"  can  now  be  drawn  by  similarity  with  the  frame  I,  since  the  members  in  the 
two  figures  are  respectively  perpendicular. 

In  like  manner  diagram  II  may  be  completed  by  transferring  c'n  from  diagram  I 
and  drawing  nc"  perpendicular  to  c'n,  thus  determining  c".  As  a  check,  the  displace- 
ments c'c"  must  be  equal  and  parallel  in  the  two  diagrams. 

(4)  A  cantilever  bridge,  similar  in  principle  to  the  Memphis  bridge,  is  represented 
in  Fig.  34D. 

It  is  assumed,  as  in  the  two  preceding  problems,  that  separate  displacement 
diagrams  have  been  made  for  each  of  the  elastic  frames,  I,  II,  III,  and  IV,  and  of  these, 
the  diagrams  for  I  and  III  can  be  completely  solved,  as  was  done  with  the  example  in 
Fig.  34A,  since  each  is  a  determinate  framed  structure  on  two  supports. 

It  is  here  deemed  necessary  only  to  complete  the  rotation  diagrams  of  the  frames 
II  and  IV.  The  displacements  of  these  frames  depend  on  those  of  the  points  c  and  m 
and  upon  that  of  the  point  s,  respectively. 

Let  it  be  assumed  that  the  points  d'  and  n'  have  been  determined  for  the  elastic 
frame  II  as  in  diagram  II,  and  let  it  be  required  to  find  the  figure  d"n"  which  shall  be 
similar  to  frame  II. 

The  members  cd  and  ran  are  elongated  by  Jl  and  J2  respectively,  and  their  elonga- 
tions must  be  applied  to  the^points  d'  and  n'  in  diagram  II,  giving  the  points  c'  and  m'. 
Now  if  the  displacements  ~c'c"  and  m'm"  be  drawn  as  found  by  diagrams  I  and  III  (not 
shown),  the  points  c"  and  ra"  are  obtained.  These  represent  the  original  positions  of 
the  points  c  and  ra.  The  point  n"  must  have  originally  occupied  the  same  relative 
level  with  ra",  also  d"  must  have  been_at  the  same  relative  level  with  c".  Lastly,  the 
line  <Fn"  must  be  perpendicular  to  dn.  Hence,  if  the  distance  x,  or  the  horizontal 
distance  between  d'  and  d",  can  be  found,  then  the  figure  d"n"  becomes  determinable 
and  the  diagram  can  be  completed.  If  it  is  granted  that  the  point  w  will  always  be 
midway  between  the  points  c  and  ra,  this  problem  becomes  definite,  and  the  value  x 


98 


KIXETIC  THEORY   OF   ENGINEERING  STRUCTURES 


CHAP.  VI 


may  be  found  from  the  figure,  thus:  x=^(e~ g +f) ,  which  is  (+)  when  measured  to 
the  right. 

Now  assuming  the  displacement  diagram  of  the  frame  IV  drawn,  and  the  points 
v'  and  t'  determined,  as  shown  in  diagram  IV,  required  to  find  the  figure  v"t" . 

The  point  v  is  a  fixed  support,  hence  v"  must  coincide  with  v' ;  also  the  direction  of 
v"t"  must  be  perpendicular  to  vt.  To  find  t",  apply  J3  from  t'  to  s',  and  then  transfer 
the  displacement  sV  from  diagram  III  (not  shown)  into  diagram  IV,  thus  giving  the 


IT 


FIG.  34o. 


original  position  of  s".  The  point  t"  must  have  dropped  from  the  height  s",  and  must, 
therefore,  be  on  the  same  horizontal  line  with  s".  Hence,  t"  is  at  the  intersection  of 
tf't"  and  the  horizontal  through  s",  and  the  figure  v"t",  similar  to  the  frame  IV,  can 
then  be  drawn. 

NOTE.     For  other  problems  see  paper  on  "The  Graphical   Solution  of  the  Distortion  of  a  Framed 
Structure"  by  the  Author  in  the  Journal  Ass'n.  Eng.  Societies,  June,  1894. 


CHAPTER   VII 

DEFLECTION   POLYGONS  OF   STATICALLY  DETERMINATE  STRUCTURES  BY 

ANALYTICS  AND  GRAPHICS 

ART.  35.     INTRODUCTORY 

The  deflection  polygon  for  any  series  of  panel  points,  either  of  the  top  or  bottom 
.chord,  may  be  found  from  a  Williot-Mohr  diagram  as  shown  in  the  previous  chapter. 

However,  when  the  deflections,  so  obtained,  are  intended  to  form  the  basis  for  stress 
computations,  then  greater  accuracy  is  actually  required  than  that  obtainable  from  the 
above  method  of  solution.  This  is  especially  true  of  large  structures,  because  the  error 
committed  by  substituting  the  tangent  of  an  arc  for  the  arc  itself  is  of  a  cumulative 
nature,  always  giving  excessive  values.  This  fact  has  been  overlooked  by  most  authors 
treating  of  indeterminate  structures. 

Professor  Mohr,  in  1S75,  established  a  very  important  relation  between  deflection 
polygons  and  equilibrium  polygons  drawn  for  simultaneous  cases  of  loading.  Accordingly 
a  deflection  polygon  may  be  derived  from  a  moment  diagram  by  dividing  the  ordinates 
of  the  latter  by  a  certain  constant  which  depends  on  the  geometric  shape  of  the  structure. 

The  elastic  deformation  in  the  angle  included  between  two  successive  chord  members 
can  always  be  expressed  in  terms  of  the  coexisting  changes  in  the  lengths  of  the  truss 
members  or  in  terms  of  the  stresses  in  these  members  for  any  simultaneous  case  of 
loading.  Such  an  angular  change  may  be  regarded  as  an  elastic  load  w,  which  designation 
will  always  be  used  in  the  following  chapters. 

The  several  elastic  loads  w  for  all  pin  points  of  an  entire  chord  will  then  suffice 
to  determine  the  deflection  polygon  for  that  chord  precisely  as  the  external  loads  P  will 
determine  an  equilibrium  polygon  from  which  moments  and  stresses  in  the  chords  may 
be  found. 

Naturally  such  a  deflection  polygon  will  furnish  displacements  in  one  fixed  direction 
only,  and  this  is  usually  all  that  is  required.  However,  two  such  polygons,  for  horizontal 
and  vertical  deflections,  respectively,  would,  if  drawn  for  the  same  pin  points  and  same 
loading,  determine  actual  displacements. 

Since  the  total  deflection  of  a  truss  is  almost  entirely  due  to  the  deformation  in  the 
chords,  it  is  often  admissible  to  neglect  the  web  system.  For  arches  with  parallel  chords 
this  is  always  true. 

The  deflection  polygon  is  usually  drawn  for  the  loaded  chord  only.  If  the  loaded 
chord  is  straight,  as  is  often  the  case,  then  the  computations  are  greatly  simplified. 

In  the  following,  two  methods  will  be  given  for  finding  such  deflection  polygons. 
In  the  first  method  the  elastic  loads  w  are  derived  from  the  changes  Al  of  all  members 

99 


100  KINETIC  THEORY  OF  ENGINEERING  STRUCTURES        CHAP,  vn 

of  the  structure,  according  to  Professor  Mohr.  The  second  method  is  a  modification  of 
the  first,  given  by  Professor  Land,  in  1887.  and  expresses  the  w  loads  in  terms  of  the  unit 
stresses  of  the  several  truss  members. 

ART.  36.     FIRST   METHOD,    ACCORDING   TO   PROFESSOR   MOHR 

(a)  Influence  on  the  deflections  due  to  chord  members.  For  any  particular  case  of 
loading,  the  stresses  and  changes  in  the  lengths  of  the  several  members  are  supposed 
to  be  known.  Neglecting  the  web  members  for  the  present,  by  regarding  their  changes 
Jl  all  zero,  the  influence  of  a  single  chord  member  is  first  considered.  See  Fig.  36A. 

The  deflection  ym  of  any  lower  chord  point  m,  due  to  a  shortening  in  the  upper 
chord  member  opposite  the  point  m,  may  be  found  from  Mohr's  work  equation.  Let 
a  unit  force  act  downward  from  m  (if  vertical  deflections  are  desired)  .  Then  from 
Eq.  (GA),  l-ym^Sma^lm,  when  there  are  no  abutment  displacements. 

Let  Mma  represent  the  static  moment  about  the  point  m  when  only  the  unit  load 
at  m  is  acting.  This  is  easily  found  from  either  one  of  the  reactions,  resulting  from 
the  unit  load,  multiplied  by  the  distance  from  m  to  that  reaction.  Then  the  stress 
Sma=Mma/rm  as  may  be  seen  from  Fig.  36A. 

Now  if  \-ym  be  regarded  as  a  moment  Mmw  due  to  a  load  w  hung  at  the  point  m, 
then  by  proportion 

Wm       Mmw  l-y 

~=  Wm  = 


S 

^ 


The  load  w,  thus  expressed,  will  be  known  as  an  elastic  load. 

Hence,  a  moment  diagram  drawn  for  the  elastic  load  wm  acting  at  m,  will  give  the 
ordinate  Mmw=ym  under  the  point  m.  Laying  off  wm  on  the  load  line  to  the  right  and 
with  a  pole  distance  H=-l,  to  scale  of  forces,  drawing  the  moment  diagram  A'm'B', 
then  the  ordinate  ym  represents  the  deflection  of  the  point  m  due  to  a  change  Mm  in  the 
length  of  the  member  fg  opposite  the  point  m,  because 

M,  =  1  •  *™  =SmMm. 


mu, 


Now  the  polygon  A'm'B'  is  also  the  influence  line  of  deflections  for  the  point  m  for 
any  moving  load  wm.  Hence,  the  deflection  ynm,  at  the  point  n,  is  that  due  to  a  load 
wm  at  the  point  m,  and  so  on  for  any  pin  point  of  the  chords. 

In  like  manner  the  deflection  yn  is  obtained  for  a  load  wn=Aln/rn  and  the  polygon 
A'n'B'  is  then  the  influence  line  of  deflections  for  the  point  n  for  a  moving  load  wn. 
The  ordinate  ymri}  under  the  point  m,  is  now  the  deflection  at  m  for  a  load  wn  at  n. 
But,  by  Maxwell's  law  y^n  =  ynm. 

By  drawing  these  influence  lines  successively  for  all  the  pin  points  of  the  "top  and 
bottom  chords,  the  several  partial  effects  of  all  the  elastic  loads  w  on  any  particular 
point  as  m  may  be  found,  and  the  summation  of  these  effects  2?/mn  will  be  the  total 
deflection  dm  for  the  point  m  resulting  from  the  given  case  of  actual  loading. 

The  summation,  however,  is  more  easily  performed  by  combining  the  several  loads 
w  into  a  load  line  21  w  and  with  a  pole  #•=!,  drawing  the  force  and  equilibrium  polygons. 


ART.  36      DEFLECTION   POLYGONS   BY   ANALYTICS  AND   GRAPHICS 


101 


The  latter  is  a  moment  polygon  and  also  a  summation  deflection  polygon,  any  ordinate 
dm  of  which  represents  the  actual  deflection  of  the  corresponding  pin  point  m  for  that 
particular  case  of  loading  and  is  measured  to  the  scale  of  lengths.  This  does  not 
include  the  effect  of  the  web  members. 

If  the  external  loads  are  applied  to  the  lower  chord  pin  points  then  the  true  deflec- 
tion polygon  for  the  lower  chord  is  the  inscribed  dotted  polygon  A'c"m"n"B' ',  because 
the  deflection  polygon  between  the  successive  floor  beams  must  be  a  straight  line. 

When  the  loads  w  are  multiplied  by  E  and  the  scale  of  lengths  is  chosen  1 :  a,  then 
by  making  the  pole  distance  H  =E  to  the  scale  of  forces,  the  deflections  will  have  the 
scale  1 :  a.  When  the  deflections  are. desired  TO  times  actual  to  the  scale  I:  a,  then  the 
pole  should  be  chosen  E/m  to  the  scale  of  forces. 


FIG.  36A. 

As  is  seen  from  the  above,  the  elastic  loads  w  are  ratios;  that  is,  they  have  no 
dimension  but  depend  on  the  unit  stresses  in  the  members  and  the  geometric  shape  of 
the  structure.  From  this  and  the  real  value  w  =  M/r,  it  follows  that  these  elastic  loads 
may  also  be  defined  as  tangents  of  the  angular  changes  in  the  angles  6,  or  simply 
wm=J6m  from  Fig.  36A. 

For  a  top  chord  point,  the  angular  change  Ad  is  positive,  while  for  a  lower  chord 
point  it  is  negative,  but  in  either  case  the  angular  change  produces  downward  deflection. 

For  a  top  chord  point 

M  .  Al  _  .  Mp 


and 


'= — .       hence 


and  for  a  bottom  chord  point 
-M=S        and 


S=- 


M 


hence  again 


Mp 


102 


KINETIC  THEORY  OF  ENGINEERING  STRUCTURES        CHAP.VII 


Therefore,  for  any  chord  point  of  a  simple  truss 


^Jlm  =       Mmpm=          M, 

''      rm  r  2    =    +IF; 


(36s) 


which  is  the  fundamental  equation  for  the  elastic  loads  w,  due  to  chord  members  only. 

(b)  Influence  on  the  deflections  due  to  web  members.  In  considering  the  effect  of  the 
web  members,  the  influence  of  a  single  diagonal  is  first  analyzed  for  the  case  when  the 
center  of  moments  falls  outside  the  span,  Fig.  36s.  The  stress  Sa  changes  sign  while 


V, 


to- 


FIG.  36B. 


the  moving  load  passes  through  the  panel  un,  hence  it  is  necessary  to  consider  whether 
the  top  or  bottom  chord  is  loaded  as  this  determines  the  load  divide. 

Let  mn  be  the  member  in  question  and  the  lower  chord  carries  the  roadbed.  Required 
to  find  the  deflections  of  all  the  lower  pin  points  for  any  given  system  of  loads.  See 
Fig.  36s. 

The  influence  line  for  the  deflection  of  a  lower  chord  point  is  determined  when  the 
deflection  for  that  point  is  known.  Having  this  influence  line,  all  the  other  required 
deflections  are  easily  found.  This  will  now  be  shown. 

The  deflection  yc  of  any  chord  point  c,  as  previously  found  for  a  unit  load  at  c,  is 
l-yc=SaJl  for  which  is  found  an  elastic  load  wc  =  Al/r,  wherein  I  and  r  refer  to  the 
member  mn  with  center  of  moments  at  o  for  a  section  pq. 


ART.  30       DEFLECTION   POLYGONS  BY  ANALYTICS  AND   GRAPHICS  103 

This  elastic  load  produces  reactions  Aa  and  Ba  and,  as  previously  shown,  the  moment 
Mwo  (f°r  aU  tne  f°rces  on  one  side  °f  tne  se°tion  aild  f°r  tne  center  of  moments  o)  is 
found  to  be  MWO  =SaM  =  l-yc- 

Hence  the  influence  line  for  the  moment  Mwo,  for_a  moving  load  w  =  Al/r,  may  be 
found.  By  laying  off  T/^cV7  as  an  ordinate  from  a'b'  under  the  point  c  to  any  con- 
venient scale,  and  drawing  a  line  oV7  to  intersect  the  vertical  through  o  in  the  point  o', 
the  two  limiting  lines  of  the  influence  area  are  determined.  Thus,  a'u'o'  represents  the 
influence  of  Ba  and  orbra/'f  thejnfluence  of  Aa.  Between  the  panel  points  n  and  u  the 
influence  line  is  a  straight  line  n'u'. 

Also,  for  any  other  position  of  this  elastic  load  the  same  influence  line  will  be  found. 
Hence  the  most  suitable  point  of  application  for  this  elastic  load  will  be  the  load  divide 
i'  for  the  panel  whose  section  is  under  consideration.  This  would  make  any  line  through 
the  load  divide  i'  an  influence  line  for  the  elastic  load  w.  Such  a  line  would  intersect 
the  verticals  through  the  two  adjacent  panel  points  in  points  n'  and  u'  and  the  two 
limiting  lines  ~rW  and  oV  must  be  made  to  intersect  in  the  point  o'  which  is  any  con- 
venient point  in  the  vertical  through  o.  These  two  limiting  lines  cut  off  the  closing  line 
tfV  on  the  verticals  through  A  and  B.  Then  any  load  w=M/r,  applied  at  the  load 
divide  i,  will  produce  a  zero  deflection  in  the  corresponding  point  c  and  a  zero  moment 
Mwo  about  the  point  o. 

Now  the  elastic  load  w  may  be  replaced  by  two  such  loads  wn  and  wu,  acting  at  the 
points  n  and  u,  respectively.  These  may  be  determined  graphically  by  drawing  the 
force  polygon  Fig.  36B,  corresponding  to  the  moment  diagram  a'u'n'V. 

The  analytic  method  for  finding  wn  and  wu  is  as  follows:  The  position  of  w  is 
evidently  immaterial,  because  the  resulting  loads  wn  and  wu  will  be  the  same  for  any 
position  of  w.  If  the  load  divide  is  chosen  as  the  point  of  application,  then  the  moment 
MWO=Q  and  the  sum  of  the  moments  produced  by  the  loads  wn  and  wu  must  likewise 
belero.  Hence,  the  resultant  of  the  two  components  must  pass  through  the  center 

of  moments  o. 

Also,  by  drawing  a  line  nh\\um,  Fig.  36s,  and  remembering  that  the  resultant  o 
wn  and  wu  must  equal  w,  then  by  taking  moments  about  71 

Ai  _  M  no     M       ,  no      r 

wuXun=--no       or       wu  =  -'==-      because       =•=-• 
r  r   un     '«  un       u 

wo      r      . 
Similarly  by  taking  moments  about  u,  and  noting  that  —j--  — ,  t 

jl  .  Al  uo      M  mo  _M 

wnXun=—-uo       or      «>„  =  —  .-=  =  — ••=---—. 
r  r   un      r    mh     r« 

The  two  new  elastic  loads,  which  were  substituted  for  w,  may  now  be  evaluated 
from  the  following  equations,  without  involving  the  lever  arm  r,  thus: 

Wu=*      and      wn (36c) 


104  KINETIC  THEORY  OF  ENGINEERING  STRUCTURES        CHAP.  Yii 

In  most  cases  this  simple  relation  proves  of  valuable  assistance,  because  r  is 
frequently  outside  the  limits  of  the  drawing. 

Regarding  the  signs  of  the  two  elastic  loads  ivu  and  wn  it  is  discerned  from  the  above 
that  the  signs  must  always  be  opposite  and  it  is  necessary  to  distinguish  which  of  the 
two  is  the  positive  load  and  then  call  the  other  one  negative.  The  sign  of  Al  alone 
determines  this  without  regard  to  the  sign  of  the  work  SaAl  and  Professor  Mohr  gives 
the  following  simple  rule,  which  will  always  identify  the  positive  elastic  load  w. 

Calling  all  top  chord  members  negative  and  all  bottom  chord  members  positive  and 
giving  the  proper  sign  to  Al  for  the  web  member  in  question,  then  the  positive  w  is  found  on 
that  side  of  the  section  or  panel  where  the  sign  of  Al  coincides  with  that  of  the  adjacent  chord. 

The  same  result  is  obtained  from  the  force  polygon,  Fig.  36s,  drawn  for  the  three 
loads  w.  This  is  also  seen  when  the  distortion  in  the  quadrilateral  ingun,  due  to  the 
change  Al  in  the  member  mn,  is  considered.  Assuming  the  top  chord  mg  immovable, 
then  —  Al  in  mn  will  cause  the  point  u  to  drop,  hence  wu  is  positive  when  the  bottom 
chord  is  the  loaded  chord.  Were  the  top  chord  the  loaded  chord,  then  nn  would  be 
considered  immovable  in  applying  this  reasoning. 

When  the  center  of  moments  o  falls  inside  the  span  there  is  no  load  divide,  whence 
the  deflection  influence  line  for  a  web  member  can  have  only  a  positive  area.  The 
maximum  stress  is  then  due  to  the  total  span  loaded,  the  same  as  for  maximum  moments. 

However,  this  in  no  wise  affects  the  previous  discussion,  but  some  proof  of  the 
identity  of  the  effects  of  the  substitute  loads  wu  and  wn  with  that  of  w  must  be  furnished 
for  this  case. 

Thus  the  elastic  load  w=Al/r,  acting  at  c,  is  in  equilibrium  with  the  reactions  A 
and  B  resulting  from  w,  and  it  is  also  the  resultant  of  the  two  substitute  loads  wu  and 
wn.  Hence  the  sum  of  the  moments  of  the  forces,  A,  B,  wu  and  wn  about  any  point  o 
in  the  plane  of  forces,  must  be  zero.  But  the  resultant  w  of  wu  and  wn,  always  passes 
through  the  center  of  moments  o,  hence  this  resultant  must  be  equal  and  opposite  to 
the  resultant  of  A  and  B.  The  method  is  thus  general  and  applies  equally  to  cases  of 
o  inside  and  outside  the  span. 

The  elastic  loads  w.u  and  wn  may  also  be  expressed  in  terms  of  the  moments  of  the 
external  forces,  as  was  previously  done  for  the  chords,  and  the  following  general  expres- 
sions are  thus  obtained: 


Al        Mp  Al        Mp 

wu=  —  =  ±—         and       wn=  —  =  =F--; 


.     .     .     .     .  .   .     (36n) 

U  ^  rn         rnr  '~  '* 

also 

Al      ,  Al    Al     Mp        .  1,11 

w~=±-*-  =  -f>       hence       -=±-T-  ......     (36E) 

•  v      'n        '  >  it       rn 

This  condition  is  easily  seen  directly  from  Fig.  36B. 

(c)  The  deflection  polygon  for  the  loaded  chord.  This  deflection  polygon  may  be 
found  by  applying  the  formulae  (36B)  and  (36o)  to  any  given  frame.  The  elastic  loads 
w  are  now  computed  for  each  chord  and  each  web  member  and  all  those  acting  at  the 
same  pin  point  are  algebraically  added  for  total  effects. 

The  deflection  polygon  is  then  found  by  combining  the  loads  w  into  a  force  polygon 
and  drawing  the  moment  or  equilibrium  polygon  exactly  as  shown  in  Fig.  36A,  following 


ART.  36      DEFLECTION  POLYGONS  BY   ANALYTICS   AND  GRAPHICS  105 

the  description  relating  thereto.     These  moments  may  also  be  computed  to  obtain  the 
deflections  analytically. 

Before  concluding  this  subject  a  few  special  cases  will  be  illustrated. 

(1)  When  there  is  no  end  post,  as  in  Fig.  36B,  then  Hitter's  section  cuts  only  two 
members  'ad  and  ac  and  some  doubt  may  exist  regarding  the  application  of  the  formula 

for  w. 

To  overcome  this  difficulty,  add  a  member  ak\\cd  and  of  any_convement  length. 
Then  the  center  of  moments  ol  will  be  the_center  for  the  member  ad  treated  as  a  web 
member.  The  two  imaginary  members  ak  and  kd  are  assumed  rigid  and  Eqs.  (36D> 
again  apply.  The  load  w  thus  found  for  the  point  a  will  have  no  effect,  and  the  result 
is  the  same  as  would  be  found  by  applying  Eq.  (36n)  to  the  center  of  moments  at  c 

with  the  lever  r\. 

(2)  When  there  is  a  vertical  end  post  and  the  deflection  polygon  for  the  lower  chord 
is  required,  then  a  load  applied  at  a,  Fig.  36c,  will  not  produce  stress  in  the  verticals 
nor  end  post.     The  end  chord  section,  and  these  members,  therefore,  do  not  influence 
the  deflection  polygon  of  the  lower  chord. 


t\ 


FIG.  36c. 

If  the  load  is  applied  at  the  upper  point  c,  then  it  is  best  to  construct  the  deflection 
polvgon  by  regarding  the  Al=Q  for  the  two  end  posts.  Now  since  the  whole  top  chord 
is  lowered  by  amounts  -M  and  -M,  of  these  vertical  end  posts,  it  is  merely  necessary 
to  raise  the  closing  line  of  the  deflection  polygon  by  these  amounts  at  the  two  respective 
ends.  The  deflection  ordinates  are  thus  corrected  by  the  addition  of  the  end  pos 

(3)  Frames  with  vertical  posts,  as  in  Figs.  36c  and  36D. 

When  the  center  of  moments  cannot  be  ascertained  by  a  section  through  three 
members,  as  for  the  member  w',  Fig.  36D,  the  elastic  load  for  the  point  a'  cannot 

directly  found.  ,, 

To  overcome  this  difficulty  assume  the  disposition  in  Fig.  36E,  where  the  small 

chord  dl  is  interposed  at  a'  and  the  vertical  aa'  is  split  into  two  posts  such  that  the 

deflection  remains  the  same  as  for  the  original  case.     Then  the  original  solution  is  again 

applicable  by  passing  two  sections  as  indicated  and  cutting  three  members  by  each  section 

The  vertical  post  is  now  made  up  of  two  substitute  posts  for  which  the  work  mu* 

be  equal  to  that  of  the  original  post,  thus: 

SI  \ 

4-2-1 

^2l    EF    I 


10G 


KINETIC  THEORY  OF   ENGINEERING  STRUCTURES        CHAP,  vn 


Also  for  each  substitute  there  will  be  a  load  w  active  at  the  center  of   moments  o 
and  the  resultant  of  these  two  loads  will  be  wa,  Fig.  860. 

Hence, 

Jl     Al 


In  the  ordinary  case,  Fig.  36c,  the  deflection  polygons  for  top  and  bottom  chords 
are  alike  when  the  web  system  is  neglected.  When  the  latter  is  included  it  is  best  to 
construct  the  deflection  polygon  of  the  chord  which  ends  in  the  supports  (the  lower 
chord  in  this  instance)  and  the  deflection  of  the  other  chord  is  then  found  by  simply 
correcting  the  first  polygon  by  the  ±  Al  of  the  verticals,  being  careful  to  regard  the  sign. 

(4)  Trusses  with  parallel  chords  become  very  much  simplified  because  the  height 
of  the  truss  and  the  inclination  of  the  web  members  are  then  uniform,  Fig.  36F. 


FIG.  36F. 

Hence  for  all  chord  members,  Eq.  (36s)  gives 

Jl 


(36F) 


For  the  web  members,  the  center  of  moments  is  at  infinity,  making  the  elastic  loads 
equal  and  opposite  for  the  same  section.  Each  pair  thus  constitutes  a  moment  which 
is  balanced  by  an  infinitesimally  small  force  acting  at  an  infinite  distance.  For  this 
and  other  reasons  it  is  better  to  determine  the  elastic  loads  for  the  web  members  directly 
from  the  shear.  Calling  the  shear  Q,  Eq.  (36o)  then  gives 


(36o) 


Here  ru=rn  for  each  panel  and  <f>  is  the  angle  of  the  diagonal  with  the  vertical. 
The  figure  shows  the  case  where  the  w  loads  are  desired  for  the  bottom  chord  panel  points. 

When  <j6  is  constant  for  all  web  members  then  the  factor  l/Er  becomes  constant 
and  may  be  applied  to  the  scale  of  the  elastic  loads  w,  which  are  then  l/Er  times  natural 
size.  Thus  by  making  the  pole  distance  Er/l  times  larger,  the  resulting  deflection 
polygon  for  web  members  will  be  unchanged. 

(5)  Composite  structures,  like  three-hinged  arches  and  cantilever  systems,  may 
likewise  be  treated  by  the  above  method. 


±111 

Al 

SI 

Ql          ] 

Wu 

ru 

EFru 

EFru  cos  (f) 

Jl 

SI 

Ql 

T  wn 

r-n 

EFrn 

EFrn  cos  <£ 

ART.  87      DEFLECTION   POLYGONS   BY  ANALYTICS   AND   GRAPHICS 


107 


For  any  positon  of  the  moving  elastic  load  w  and  any  panel,  the  same  work  equation 
applies  and  also  Eq.  (36e) . 

Hence,  with  due  regard  to  signs,  the  influence  area  of  a  deflection  y  is  again  the 
influence  area  for  the  moment  M^  and  also  the  influence  area  of  the  particular  member 
under  consideration.  It  is  only  necessary  to  observe  the  signs  of  the  loads  w  and  the 
corresponding  position  of  the  closing  line  of  the  deflection  polygon. 

The  w  loads  which  result  from  the  chord  members  will  have  the  same  sign  as  the 
moment  influence  line  for  that  panel.  When  the  moment  changes  sign  in  any  par- 
ticular panel,  as  might  be  the  case  in  cantilever  systems,  then  the  w  loads  of  the  adjacent 
pin  points  must  be  of  opposite  signs. 

The  w  load  resulting  from  the  diagonals  will  always  be  of  opposite  signs  according 
to  the  rule  above  given,  no  matter  whether  the  center  of  moments  for  such  diagonals 
falls  inside  or  outside  the  span. 

(6)  For  indeterminate  systems  the  foregoing  method  is  made  applicable  by  removing 
the  redundant  members  or  reaction  conditions  and  replacing  these  by  external  forces 
X  acting  on  the  principal  system  in  the  manner  described  in  Art.  7. 

The  deflection  of  an  indeterminate  system  under  certain  loads  P  is  the  same  as  the 
deflection  of  the  statically  determinate  frame  subjected  to  the  loads  P  and  X  and  the 
work  Eq.  (GA)  is  in  like  manner  applicable  to  this  principal  system,  so  loaded. 

An  example  showing  the  application  of  the  above  method  may  be  found  in  Table  50s. 


ART.  37.     SECOND   METHOD,   ACCORDING   TO   PROFESSOR   LAND 

(a)  The  w  loads  in  terms  of  the  unit  stresses  in  the  members. 

This  method  is  sometimes  preferable  to  the  former  when  the  deflection  polygon  of 
any  succession  of  members  is  desired  without  regard  to  the  remaining  members  of  the 
structure.  The  first  method  is,  however,  preferable  when  a  more  general  solution  is 
required. 

Here,  any  succession  of  members  or  rods  is  treated  purely  as  a      kinematic  chain  " 
involving    only    the    angular    changes     occurring 
in      the     angles      included      between      successive 
members  and  the  changes  Al  in  the  lengths  of  these 
members. 

The  problem  resolves  itself  into  two  parts: 
(a)  to  find  the  changes  in  the  angles  of  any 
triangle  resulting  from  changes  M  in  the  lengths 
of  the  sides,  and  (6)  to  evaluate  the  loads  w  in 
terms  of  these  changes  and  finally  to  construct  the 
deflection  polygon. 

(b)  To  find  the  changes  in  the  angles  of  a 
triangle. 

Let  a,  /?  and-  7-  be  the  three  angles  of  a  triangle  ABC  and/a,/6  and/c  the  unit  stresses 
respectively  in  the  three  members  opposite  these  angles,  as  given  in  Fig.  37A.  Also  let 


108  KINETIC  THEORY  OF  ENGINEERING   STRUCTURES        CHAP.  VII 

Ja,  AB  and  Af  be  the  changes  in  the  three  angles  resulting  from  the  changes  Jla,  Jlb 
and  Alc  in  the  lengths  of  the  three  sides. 

Then  from   the   figure,   sin  a\  =x/lc   and   sin  «2  =y/lb,    and   by   differentiation   and 
treating  all  quantities  as  variables, 


cos 


lcAx—xAlc  lbJy-yJlb 

and       cos  a^Ja<>=—  —  -. 

r  *  /«.* 


Adding  these  expressions  and  solving  for  Aa\  +  J«2  =  J«  gives 

Jx—x 


Jlb 


/  7 

?- (37A) 


But  Jc  cos  «i  =£{,  cos  «2  =/",  hence  Eq.  (3?A)  becomes 

..  (37B) 


ry         b 
Now  from  Fig.  3?A, 

Jx_Jy_f»f       Mc_fc.       Mb_fb.      ^_cot/?.       and       y._cotr 
T~Y    FJ        Zc  ~£'        Z6~£'       r~     *<*>  r~      tr' 

which  values  substituted  into  Eq.  (3?B)  give 

^a  =  (/a-/6)cotr  +  (/a-/c)cot^.   .  _.    ,  •"."-  \     .     .     (37c) 

The  values  EJ{3  and  EAj-  can  be  derived  in  the  same  way  to  obtain  the  following: 
E  Aa  -  (fa  -fb)  cot  r  -  (fc  -fa)  cot  B  1 

=  (fb-fc]  cota-(/a-/6)  cotr  \.    .    ."../.     .     (37n) 
=  (fe-fJ  cot  B-(fh-ft  cot  a 


(3?E) 


wherein  tensile  stresses  /  are  positive  and  compressive  stresses  are  negative. 

Since  the  sum  of  the  three  angles  of  a  triangle  is  always  constant,  therefore, 


whence  the  third  value  is  easily  found  after  having  computed  any  two  from  Eqs.  (37n). 

The  above  formula  are  extensively  used  in  Art.  61  dealing  with  secondary  stresses. 

When  temperature  changes  are  to  be  included  then  the  unit  stresses  /  must  be 
corrected  by  dE.  A  positive  t  produces  elongation  in  all  members,  hence  dE  must 
then  be  positively  applied  to  all  values/.  The  contrary  is  true  for  —t. 

The  change  in  any  peripheral  angle  ^  of  a  frame  is  easily  found  by  summing  the 
changes  Aa  occurring  in  the  several  angles  a  whose  apices  meet  in  that  angle  0.  Then 
^  =  Sa  and  J^  =  2Ja,  making 

EI>Aa       ......     ....     (37F) 


ART.  37      DEFLECTION  POLYGONS   BY  ANALYTICS  AND   GRAPHICS 


109 


using  the  minus  sign  when  0=360°  — Ha  which  is  the  case  when  the  vertices  of  the 
angles  a  are  on  the  opposite  side  of  the  kinematic  chain  from  the  peripheral  angle  <p. 
The  values  EJa  are  obtained  from  Eqs.  (3?D). 

The  products  of  the  form  (fa—fb)  cot  7-  in  Eq.  (37c)  may  be  graphically  found 
from  a  large  scale  truss  diagram  asjndicated  in  Fig.  3?A.  The  quantity  (fa—fb)  is  laid 
off  perpendicular  to  either  AC  or  BC  so  as  to  include  the  angle  ?.  The  base,  to  scale 
of  forces  /,'  will  then  be  the  desired  quantity.  The  sign  is  easily  found  by  inspection 
of  the  given  data  and  depends  on  the  sign  of  (fa  —fb)  and  the  sign  of  the  cot  f. 

(c)  To  evaluate  the  elastic  loads  w  in  terms  of  the  angle  changes  J  0  and  the  changes 
M  in  the  lengths  of  the  members,  and  finally  to  construct  the  deflection  polygon  for  the 
kinematic  chain. 

In  Fig.  37  R  such  a  chain  of  chord  members  is  shown  and  the  J</»  and  Al  are  now 
supposed  to  be  given.  The  system  is  referred  to  coordinate  axes  (x,  y). 


A' 


FIG.  37s. 

Now  let  /l=the  angle  which  any  member  makes  with  a  line  parallel  to  the  x  axis 
and  through  the  right-hand  end  of  the  member.  See  Fig.  37s  regarding  the  sign  of  A 
for  different  members. 

OA,  81,  32  are  the  displacements  of  the  points  A,  1,  2,  etc.,  respectively,  measured 
from  the  closing  line  A'B'  and  parallel  to  the  given  y  axis. 

w\,  w2,  w3   are  the  elastic  loads,  the  moment  polygon  for  which  represents  the 

'  These  loads  are  now  to  be  found 

The 


deflection  polygon  A'l'2'3'  with  closing  line  A'B'. 

and  then  the  polygon  is  easily  constructed  as  for  the  first  method  above  given, 
loads  w,  parallel  to  the  y  axis,  are  applied,  as  before,  in  the  direction  in  which  the 
deflections  are  to  be  measured. 

In  the  figure   rF|| 'ATB7\\Wc,  hence  it  follows  that  ce^(o2— diK-  and  ac  =  d2—  d3 
and  by  addition  of  these  equations,  ae  is  obtained  as 


)-T  +  (^2— 03)- 
d2 


(37o) 


1 10  KINETIC  THEORY  OF  ENGINEERING  STRUCTURES        CHAP.  VII 

But  since  the  sides  of_the  deflection  polygon  are  respectively    parallel  to  the  rays 

y-f  f>  nij  j.  

of  the  force  polygon,  then  -r-=-fr.     This  value  of  ae  inserted  in  Eq.  (37c)  gives 

ds     ti 


Also  from  Fig.  37s, 

.Vi  — 2/2  =fe  sin  4      •     • :. 

The  differential  of  Eq.  (37j)  gives 

Ji/i  —  Ay2  =A12  sin  X2  +12  cos  k2d&2  =§i  —d2        (37iv) 

which  must  equal  o\  —  o2  because  'both  expressions  represent  the  deflection  of  the  point 
2  with  respect  to  the  point  1. 

Also  from  the  figure  d2  =12  cos  X2  which  divided  into  Eq.  (37K)  gives 


and  in  like  manner  is  found  for  +  ^ 


—  , 

--  -j  —  =-;-  tan 
"3          h 

Making  H  =  1  and  substituting  these  values  into  Eqs.  (37n)  then 


—  ~  tan  A2+-r-^  tan  A3  .......     (37L) 


But  180°->(2  +  >l3  =    2  then  - 


Also  -r^=Tf       and       -7-^=^,       whence  Eq.  (37i.)  becomes 
12      &  /a      Jb 


f2t&n  A2+/3tan  4     .......     (3?M) 

The  general  expression  for  any  pin  point  n  is  then 

Ewn  =  ±E4  tyn  -fn  tan  Xn  +/„  +  1  tan  4  .  +  1         .....     (3?N) 


in  which  Ed<f>n  is  given  by  Eq.  (37r)  for  top  chords,  using  the  —  sign  for  bottom  chords. 

For  any  negative  angle  ^  (see  Fig.  37s)  tan  A  also  becomes  negative.  Similarly 
the  signs  of  the  unit  stresses  /  must  be  considered  in  substituting  values  in  Eqs.  (37r) 
and  (37N).  The  equations  as  they  stand  are  written  for  positive  values  in  all  cases. 

//  the  elastic  loads  w  are  multiplied  by  E  and  the  force  polygon  be  constructed  to  any 


ART.  37      DEFLECTION   POLYGONS  BY    ANALYTICS   AND  GRAPHICS  111 

convenient  scale  of  forces  with  a  pole  H=E  to  scale  of  forces,  when  the  scale  of  the  truss 
diagram  is  l:a,  then  the  deflections  will  be  actual  to  the  scale  I:  a.  But  if  the  deflections 
should  appear  m  times  actual  when  measured  with  the  scale  1  :  a,  then  the  pole  distance  must 
be  made  E/m  to  the  scale  of  forces. 

Finally  the  closing  line  A  'B'  is  found  from  the  reaction  displacements.  If  these 
are  zero,  then  dA  and  dB  are  zero  and  the  closing  line  must  join  the  ends  of  the  equilibrium 
polygon.  Otherwise  $A  and  dB  must  be  ascertained  from  the  conditions  of  the  supports. 

When  the  web  system  does  not  include  vertical  members  then  it  is  preferable  to 
choose  these  web  members  for  the  kinematic  chain  rather  than  a  chain  of  chords,  because 
the  deflection  polygon  will  then  furnish  the  deflections  of  all  the  pin  points  instead  of 
only  those  belonging  to  one  chord. 

When  the  deflections  in  one  direction  only  are  required,  then  the  above  polygon 
affords  a  complete  solution,  but  for  a  general  determination  of  displacements  in  the 
plane  of  the  truss,  a  second  deflection  polygon  would  be  required,  giving  deflections 
perpendicular  to  the  direction  chosen  for  the  first  polygon.  In  the  above  illustration 
this  would  be  parallel  to  the  x  axis  and  the  resultant  displacement  of  any  point  would 
be  the  resultant  of  the  horizontal  and  vertical  displacements  of  that  point. 

The  displacements  in  the  x  direction  could  be  computed  from  the  same  Eq.  (37isr) 
by  substituting  —  cot  X  for  +tan  A  in  each  case,  but  a  better  solution  is  given  below. 

The  method  fails  when  any  angle  ^=90°,  hence  no  member  of  the  kinematic  chain 
should  be  parallel  to  the  direction  of  the  deflections.  If  this  occurs  then  a  pair  of 
substitute  members  must  be  employed  as  was  done  in  Fig.  36B.  For  this  reason  also, 
the  w  loads  from  Eq.  (37x)  cannot  usually  be  employed  for  finding  horizontal  displace- 
ments as  per  Art.  38. 

For  a  straight  chord,  the  Eqs.  (37F)  and  (37*)  give  the  following  simple  formula 
for  wn 

.........     (37o) 


Example.  The  above  method  will  now  be  illustrated  by  the  following  example 
of  a  two-hinged  framed  arch  as  shown  in  Fig.  37c.  The  same  problem  is  also  solved 
in  Art.  50,  using  the  method  given  in  Art.  36. 

The  unit  stresses  and  angle  functions  appear  in  the  truss  diagram  of  Fig.  37c  arid 
the  nomenclature  used  in  Eqs.  (37o)  as  indicated  on  a  small  triangle,  is  successively 
applied  to  all  the  triangles  of  the  frame. 

The  deflection  polygons  are  drawn  first  for  the  bottom  chord  and  then  for  the  top 
chord,  both  for  the  conventional  loading  Xa=l,  and  the  computations  of  the  Ew  loads 
are  given  in  detail  for  the  bottom  chord  only,  because  these  represent  the  general  case 
necessitating  the  use  of  Eqs.  (37D),  Eq.  (37r)  and  Eq.  (3?N).  For  the  top  chord,  the 
computations  do  not  require  the  use  of  Eq.  (3?N)  and  the  resulting  values  of  the  loads 
Ew  only  are  given  without  the  details  of  the  figuring. 

In  the  present  problem,  the  method  would  not  apply  to  the  kinematic  chain 
f/0^1^1^2^2,  etc.,  because  the  verticals  are  parallel  to  the  direction  of  the  vertical 
deflections.  Hence,  the  chords  are  treated  separately,  thus  avoiding  all  vertical  members 
in  the  chains. 


112 


KINETIC   THEORY  OF   ENGINEERING  STRUCTURES        CHAP.  VI] 


15* 


15' 
U0    fg-O.OI07 


3     The  ^  ig"r«»  "bo'v*  +he  members  reprlcserrl  unit  SfreSSesfl 

Kip. 


BOTTOM  CHORD  DEFLECTION  POLYGON 


All  deflections  af*£-  times  actual  'in  f>et,  when  meaeUred  + 

scale  of  lengths .  The  figures  in  ft.  'are  E  time*  actual 

I  I 

I  I 

I  I 

I 
I 
I 


LENGTHS' 


DEFLECTIONS. 


FIG.  37c. 


AKT.  157      DEFLECTION  POLYGONS  BY   ANALYTICS   AND   GRAPHICS  113 

The  elastic  load  EwQ  for  the  bottom  chord  has  no  effect  on  the  deflection  of   this 
chord  and  is,  therefore,  neglected. 

Computation  of  the  elastic  loads  Ew  for  the  bottom  chord  panel  points. 

For  panel  point  L  i  : 

By  Eqs.  (3?D),  #J«i  =(-0.0257  -0.0503)3.059  -       (0.0255+0.0257)0.454  =    -0.2557 

=0.0  (0.0503+0.0107)0.890  =    -0.0543 

=   (0.0625-0.0347)0.1125  -(-0.0342-0.0625)1.394  ==+0.1379 


ByEq.  (37F),    -EAfa  =EI>Aa  =  -0.1721 
By  Eq.  (37N),      Ewi  =0.1721  -0.0255x0.660+0.0347X0.5596=  +0.175 

For  panel  point  L2: 

By  Eqs.  (37D),  EActi  =(-0.0342-0.0625)1.394  -  (0.0347+0.0342)0.5596=  -0.1733 

EAa%  =0.0  (0.0625  +0.0352)0.7173  =  -0.0701 

EJci3  =  (0.0791-0.0493)0.4192  -(-0.0498-0.0791)0.963  ==+0.1366 


By  Eq.  (37r),    -EJfa  =EI>Aa  =  -0.1068 
By  Eq.  (37N),     Ew2  =0.1068  -0.0347X0.5596  +0.0493x0.4312=  +0.109 

For  panel  point  L:;: 

By  Eqs.  (37D>,  EActi=  (-0.0498  -0.0791)0.963  -       (0.0493+0.0498)0.431  =-0.1668 

#J«2=0.0  (0.0791+0.0736)1.038  =-0.1585 

=     (0.1065-0.0687)0.821  -(-0.0657-0.1065)0.649  ==+0.1428 


By  Eq.  (37r),    -EAfa  =EI>Aa  =  -0.1825 
By  Eq.  (37N),  Ew3  =0.1825-0.0493  X0.4312  +0.0687  X0.3147  =  +0.183 

For  panel  point  L4: 

By  Eqs.  (37o),  EAon  =(-0.0657-0.1065)0.6487-     (0.0687+0.0657)0.3147  =  -0.1540 
=0.0  (0.1065+0.1000)1.542    =-0.3184 

=     (0.0889-0.0990)1.409   -(-0.0579-0.0889)0.462    ==+0.0536 


By  Eq.  (37r),   -EA$*=ET>Aa  =  -0.4188 
ByEq.  (37N),     Ew±  =0.4188  -0.0687X0.3147  +0.0990X0.1865=  +0.416 

For  panel  point  L-  : 

By  Eqs.  (37D),  EAa^  =(-0.0579-0.0889)0.462  -       (0.0990+0.0579)0.1865=  -0.0971 

EJa2=0.0  (0.0889+0.1081)2.164   =    -0.4263 

=     (0.0541-0.1207)2.111  -(-0.0270-0.0541)0.400   •    -0.1082 


By  Eq.  (37r),   -EA^=E^Aa  =  -0.6316 
ByEq.  (37ar),     Ew5  =0.6316  -0.0990X0.  1865  +0.1207  X0.0621  =  +0.621 

For  panel  point  L,;: 

By  Eqs.  (37D),jBJ«i=  (-0.0270  -0.0541)0.400   -       (0.1207+0.0270)0.062   =-0.0416 
£JJa2==0.0  (0.0541+0.1290)2.500   =    -0.4578 


By  Eq.  (37p),   -i^J06=i^SJa=  -0.4994 
ByEq.  (37u),   ^Ewe  =0.4994  -0.1207  X0.0621  =  +0.492 


114  KINETIC  THEORY  OF   ENGINEERING  STRUCTURES        CHAP.  VII 

The  minus  sign  in  Eq.  (37r)  applies  here  because  the  angles  a  are  on  the  side  of 
the  chord  opposite  to  the  angle  <p. 

The  Ew  elastic  loads  are  now  combined  into  a  force  polygon  and  the  equilibrium 
polygon  drawn  for  these  loads  acting  through  the  several  lower  chord  panel  points  will 
then  represent  the  deflection  polygon  for  the  lower  chord. 

If  the  pole  distance  H  were  made  equal  to  E  then  the  deflections  would  be  actual 
to  the  scale  of  lengths  according  to  the  rule  above  given.  However,  this  would  be  too 
small  a  scale  for  accurate  results  and  therefore,  the  pole  distance  was  made  equal  to  4w 
units,  giving  deflections  E/4  times  actual  to  the  scale  of  lengths  of  the  drawing.  By 
constructing  a  scale  four  times  smaller  than  the  scale  of  lengths,  the  deflection  ordinates 
E  times  actual  may  be  scaled  directly  from  the  drawing,  and  these  values  are  written 
on  the  ordinates. 

The  same  problem  is  solved  by  the  graphic  method  in  Fig.  50s  and  again  by  the 
method  of  Art.  36  in  Table  50B. 

Using  the  Ew  loads  computed  for  the  top  chord  panel  points  the  top  chord  deflection 
polygon  is  obtained  in  precisely  the  same  manner  and  these  loads  and  the  deflections 
found  for  the  top  chord  are  comparable  with  the  values  in  Art.  50  just  referred  to. 

The  numerical  values  of  the  Ew  loads  for  the  top  chord  panel  points  are  as  follows: 

#w0=  -0.139  Ew2=  +0.117  Ew4=  +0.415  %Ew6=  +0.479 

Ewi  =  +0.098  Ew3  =  +0.203  Ew5  =  +0.608 

The  details  for  the  computation  of  the  load  Ew%  are  given  for  illustration.     Thus  : 

By  Eqs.  (37o),  EAcn=(  0.0625+0.0352)0.7173  -(-0.0498-0.0625)1.394=  +0.2266 
=  (  0.0493+0.0498)0.431  -(  0.0791  -0.0493)0.419=  +0.0302 
=  (-0.0657  -0.0791)0.963  -0.0  -0.1394 


By  Eqs.  (37r)  and  (37o)  Ew2  =EA$2  =#Z  Jo:  =  +0.1174 

which  follows  for  a  straight  chord  when  the  angles  a  and  ^  are  'on  the  same  side  of  the 
chord. 


ART.  38.     HORIZONTAL   DISPLACEMENTS 

To  find  the  displacements  in  a  horizontal  direction  when  the  deflection  polygon 
for  vertical  deflections  is  given. 

It  is  readily  seen,  from  the  previous  description,  that  the  elastic  loads  w  are  indepen- 
dent of  the  direction  of  the  deflections  because  in  every  instance  their  value  is  dependent 
on  the  shape  of  the  truss  and  the  changes  in  the  lengths  of  the  members,  regardless  of 
how  these  were  produced.  See  Eqs.  (36s)  and  (36c).  Hence,  the  same  elastic  loads 
may  be  employed  to  find  deflections  in  any  direction. 

The  force  polygon,  used  for  the  deflection  polygon  for  vertical  deflections,  will  thus 
be  the  same  for  any  other  direction  of  deflections  when  revolved  through  an  angle 
equal  to  the  angle  included  between  these  two  directions.  When  horizontal  and  vertical 


ART.  38      DEFLECTION  POLYGONS   BY  ANALYTICS  AND   GRAPHICS  . 


115 


deflections  are  under  consideration  this  angle  is  90°  and  the  same  force  polygon  may 
readily  serve  both  purposes,  since  the  rays  for  horizontal  deflections  are  then  perpen- 
dicular to  the  rays  drawn  for  the  vertical  deflections. 

In  Fig.  38A,  let  the  succession  of  members  A  to  B  represent  the  bottom  or  tension 
chord  of  some  structure,  fixed  at  A  and  supported  by  a  roller  bearing  on  a  horizontal 
plane  at  B.  Also  given  the  deflection  polygon  A'c'd'e'f'B'  to  find  the  deflection  polygon 
A"c"d"e"f"B"  for  horizontal  deflections.  The  loads  w  are  all  multiplied  by  E  and  the 
deflections  are  m  times  natural. 

The  left-hand  end  of  the  chord  havirig  no  horizontal  motion,  it  is  most  convenient 
to  call  all  horizontal  deflections  positive  to  the  right. 

The  pin  points  are  horizontally  projected  on  to  the  closing  line  A"e0,  which  latter 
is  perpendicular  to  the  closing  line  A'B'.  The  polygon  A"c"d"e"f"B"  is  then  drawn 


*>-dL^ 


FIG.  38A. 

by  making  the  sides  perpendicular  to  the  respective  rays  of  the  force  polygon  and 
maintaining  the  same  order  of  succession  in  the  pin  points  formerly  used  in  constructing 
the  deflection  polygon  for  vertical  deflections.  The  horizontal  abscissa  BQB"  is  then 
the  horizontal  displacement  of  the  point  B.  Similarly  the  abscissa  cfod"  is  the  horizontal 
displacement  of  the  point  d,  etc. 

For  any  horizontal  member,  as  ef,  the  horizontal  displacement  of  the  point  /  with 
respect  to  e  must  naturally  be  the  M  for  that  member.  Hence  since_the  scale  of  deflec- 
tions is  m  times  actual  to  the  scale  of  lengths  the  displacement  e"f" =mAl  to  the  scale 
of  lengths  used. 

If  the  point  B  is  made  to  roll  on  some  inclined  plane  ik  instead  of  the  horizontal, 
then  a  new  closing  line  A7%  must  be  determined  for  the  two  deflection  polygons,  as 
follows:  draw  a  line  'B^' \\~ik  and  erect  a  perpendicular  to  AB  prolonged,  and  passing 
through  B".  The  intersection  of  these  two  lines  gives  k"  and  the  vertical  ordinate  3B 
represents  the  vertical  displacement  of  the  support  B. 


116 


KINETIC  THEORY  OF  ENGINEERING   STRUCTURES        CHAP.  VII 


Hence,  making  B'k'  =dB  the  closing  line  A'k'  may  be  drawn.  Also  a  line  A"k0  J_  A'k' 
becomes  the  required  closing  line  for  the  horizontal  deflection  polygon. 

The  same  solution  will  apply  to  deflections  in  any  direction  other  than  90°  with 
the  vertical.  Also,  when  the  deflection  polygon  for  vertical  deflections  of  all  points  of 
a  truss  is  given,  as  in  Fig.  36A,  then  the  horizontal  deflections  of  all  pin  points  are  found 
precisely  as  above  by  adhering  strictly  to  the  sequence  in  which  these  points  occur  on 
the  given  deflection  polygon. 

The  change  dAB  in  the  length  of  the  chord  AB,  Fig.  38A,  may  easily  be  found  by 
taking  the  loads  w  parallel  to  AB.  In  this  case  the  displacement  d^B  becomes  the 
intercept  on  AB  produced  and  included  between  the  extreme  rays  of  the  equilibrium 
polygon.  Hence  this  displacement  is  equal  to 


,ww 


(38A) 


wherein  y  is  the  ordinate  of  any  pin  point  measured  normally  to  A B. 
This  same  result  may  also  be  found  by  computation  from 


(38u) 


which  follows  for  a  pole  distance  of  unity. 

For  the  case  given  in  Fig.  37fi,  where  the  angle  changes  A$  and  the  changes  in  the 
lengths  of  the  members  are  given,  the  total  effect  on  the  length  of  span  AB  then  becomes 
for  n  members, 

^B  =  Si"~V0  +  SrJZ  cos  (;-«),  .--..    ,•>.;-.     .     (38c) 

where  a  is  the  angle  which  the  span  AB  makes  with  the  horizontal. 

Example.  The  lengthening  d^B  for  the  span  AB  due  to  the  loading  Xa  =  l  in  Fig. 
37o  is  computed  from  Eq.  (38c) ,  using  the  values  E Al  as  given  in  Fig.  50A.  The  value 
\EdAB  was  found  to  check  the  value  determined  graphically  in  Fig.  50s.  The 
values  Ed<{>  are  those  above  computed  for  the  bottom  chord  panel  points  and  a  =0 
because  the  span  is  symmetric. 


Point. 

E4$ 

Feet. 

EyJ<l> 

E4l  Fig.  50A. 
Feet. 

cos  >l 

KJl  co*  X 

A 

0.1721 

6.188 

1.0649 

0  .  286 

0.834 

0.2385 

L, 

0.1068 

14.58 

1.5571 

0.597 

0.873 

0.5212 

L3 

0.1825 

21.05 

3.8416 

0.805 

0.918 

0.7390 

L< 

0.4188 

25.77 

10.7925 

1.080 

0.954 

1.0303 

L, 

0.6316 

28.568 

18.0435 

1.511 

0.983 

1.4853 

L, 

£  =  0.4994 

29.50 

14  .  7323 

1.814 

0.998 

1.8104 

=     50.0319 


5.8247ft. 


These  values  substituted  into  Eq.    (38c)    give  for  a  lengthening  E  times   actual, 
4B=iE27/^  +  A#2J/cos  >l=  50.03 19  +5.8247  =55.8566  ft.,  where  £-29000  kips  per 
sq.  inch. 


r.  39      DEFLECTION   POLYGONS  BY   ANALYTICS   AND  GRAPHICS  117 


ART.  39.     DEFLECTION  OF  SOLID  WEB   BEAMS 
(a)  Deflection  due  to  moments.    The  differential  equation  of  the  elastic  curve  is 

dx2=-E~r      '    '     '  •     (39A> 

|         Eq.  (36s)  gives  the  elastic  loads  w,  for  only  one  chord,  in  terms  of  moment  and 
truss  dimensions  as 

Mp      Ml 

7/1  = L  = 

r2      EFr2' 

Considering  both  chords,  each  of  area  F,  and  neglecting  the  web  effect,  which  is 
usually  quite  small,  then  for  a  unit  length  of  girder 


EFr2      El  ' 

Hence  Eq.  (39A)  gives  directly  the  elastic  loads  w  per  unit  length  of  girder  as 

d2v     M 


(39s) 


An  equilibrium  polygon  drawn  for  these  loads,  with  pole  distance  equal  to  unity, 
gives  deflections  to  the  scale  of  lengths  chosen  for  the  drawing. 

By  treating  the  moments  M  per  unit  length  of  girder,  as  loads  which  now  become 
El  times  too  large,  and  constructing  an  equilibrium  polygon  for  these  loads  M  with  a 
pole  distance  H=El,  the  same  deflection  polygon  is  again  obtained,  giving  deflections 
to  the  scale  of  lengths  of  the  drawing.  In  other  words  the  moment  polygon  for  any 
case  of  loading  becomes  the  load  line  for  the  deflection  polygon  corresponding  to  such 
case  of  loading. 

Deflections  are  usually  drawn  several  times  actual  size,  in  which  case  the  pole 
must  be  divided  by  as  many  times  the  scale  of  the  drawing.  Thus,  if  the  scale  of  lengths 
is  1  :n  and  the  deflections  should  appear  twice  actual,  then  the  pole  H  =EI/2n. 

When  the  value  /  is  variable,  the  pole  distance  is  made  to  vary  as  a  function  of  7, 
as  illustrated  in  the  example  below.  See  also  Art.  47. 

(b)  Deflection  due  to  shear.  From  Eq.  (15M)  the  deflection  of  a  straight  beam 
when  acted  on  by  shearing  forces  only,  is 

s»n        ^f) 

(39c) 


Disregarding  all  unnecessary  refinements,  since  shear  deflections  are  small  compared 
to  the  total  deflection,  tire  value  /?  may  be  taken  as  constant  and  Q  will  be  assumed 
uniformly  distributed  over  the  web  plate  of  sectional  area  FI  =^ 


118  KINETIC   THEORY  OF  ENGINEERING  STRUCTURES        CHAP.  VII 

For  a  single  concentrated  load  'dQ/'dPm  =  1  and  Q  =R  =the  end  reaction,  hence 

3        /•«    30   ,        8    /•» 


Taking  (?  =0.3332?  and  Fi=F/3   and   assuming  an  average  value  for   ,3=2.5  for| 
ordinary  girder  sections,  then  approximately 


(c)  Deflection  due  to  combined  shear  and  moments.  Ordinarily  it  will  suffice  to 
figure  the  deflection  due  to  shear  for  a  point  at  or  near  the  point  of  maximum  moments 
and  to  determine  the  percentage  which  this  dm  is  of  the  moment  deflection  ordinate 
at  the  same  point.  All  other  moment  deflection  ordinates  may  then  be  increased  in 
the  same  proportion  to  obtain  the  deflection  polygon  for  combined  shear  and  moment 
effects. 

For  a  uniformly  loaded  beam  or  constant  /  and  neglecting  shear  effect,  the  deflection 
becomes 


1    r 
°=EIJ0 


When  the  depth  of  a  girder  is  constant,  but  the  moment  of  inertia  varies  so  as  to 
maintain  a  constant  stress  on  the  extreme  fiber  at  all  sections,  then  the  ratio  M/EI 
becomes  constant  and  the  deflection  for  such  a  case  would  be 


-if: 


where  Im  is  the  moment  of  inertia  at  the  section  of  maximum  stress. 

When  the  moment  of  inertia  varies  abruptly  the  deflection  may  be  expected  to 
fall  between  the  two  above  values,  making  the  numerical  coefficient  close  to  1/9. 
Taking  in  the  shear  effect,  the  approximate  formula  for  maximum  deflection,  of  a  beam 
of  variable  /,  becomes 

%       •'"max^      |  ""*max  fon^\ 

"-*"'       '     ' 


However,  for  general  cases  of  loading  and  variable  7  the  graphic  solution,  above 
given  under  Art.  39A,  should  be  employed. 

(d)  Example.  Given  a  plate  girder  of  variable  section  and  uniformly  varying  water 
load  as  shown  in  Fig.  39A,  required  to  find  its  deflection  polygon.  The  girder  is  designed 
as  a  double  web  beam  and  normally  occupies  a  vertical  position  so  that  there  are  no 
dead  load  stresses. 

These  girders  are  spaced  9.175  ft.  center  to  center  so  that  each  will  carry  a  full 
water  load  P=31.25X9.175(?-a)2  giving  rise  to  the  end  reactions  A  and  B,  and  moments 
M,  all  as  indicated  in  the  diagrams,  Fig.  39A. 


DEFLECTION   POLYGONS  BY  ANALYTICS  AND   GRAPHICS 


119 


FIG.  39A. 


120  KINETIC  THEORY  OF  ENGINEERING  STRUCTURES        CHAP.  VII 

The  moments  of  inertia  were  computed  for  the  various  sections  on  gross  areas,  and 
are  plotted  in  the  upper  diagram. 

The  bending  moments  were  also  computed,  using  the  formula 


(39H) 


and  the  results  were  plotted  in  millions  of  foot-pounds  in  the  second  diagram. 

The  w  loads  were  then  taken  as  the  moments  M  over  certain  convenient   lengths 
Ax,  instead  of  unit  lengths  according  to  the  formula  (39s)  ,  whence 


The  lengths  Ax  are  usually  chosen  with  respect  to  the  girder  sections,  so  that  /  is 
constant  over  each  length  Ax.  Where  the  depth  of  section  is  variable  the  mean  value 
of  /  is  taken  for  each  length  Ax. 

Since  M  and  Ax  are  both  expressed  in  feet,  while  E  and  7  are  both  for  inches,  the 
factor  12X12  =  144  is  introduced  into  Eq.  (39  j)  to  establish  a  true  ratio  between  like 
units.  Also,  M  and  E  are  both  expressed  in  millions  of  pounds,  making  the  unit 
weight  one  million  pounds.  This  furnishes  the  true  values  for  w  in  terms  of  the  given 

data  as 

144MAx    MAx    MAx  ,_.   . 

-—-'        •  •  •  •  •  •  (39K) 


144 

wherein  E  =28  is  chosen  low  rather  than  high. 

Now  the  scale  of  lengths  was  taken  as   1  :  120,  and  if  the  deflections  are  to  appear 

twice  actual  size  then  the  pole  must  be  made  equal  to 

(*» 

OO  T 


The  pole  distances  are  thus  a  constant  function  of  7  and  may  be  found  for  all 
values  of  7. 

The  force  polygon  is  now  easily  drawn  to  any  convenient  scale  of  forces,  using  the 
same  scale  for  the  loads  w  and  the  pole  distances  77.  This  scale  has  no  influence  on 
the  deflection  polygon,  but  merely  affects  the  size  of  the  force  polygon. 

The  deflection  polygon  is  then  constructed  as  the  equilibrium  polygon  of  the  w 
loads  with  the  various  pole  distances,  and  the  ordi  nates  included  between  this  polygon 
and  the  closing  line  A'B'  will  represent  the  deflections  to  twice  natural  scale.  Had 
the  pole  distances  been  taken  twice  as  long,  then  the  deflections  would  have  been  actual. 

The  deflection  due  to  shear,  at  the  point  of   maximum    moments,  is  found  from 


5x537X12 
Eq.  (39n)  as  om=    „      =—  ;        —  ^-p-=0.09S  inch  which  is  10.5  per  cent  of  the  maxi- 


mum  deflection  due  to  bending  alone. 


ART.  39       DEFLECTION  POLYGONS   BY  ANALYTICS   AND  GRAPHICS  121 

The  total  deflection  of  the  girder  is  therefore  10.5  per  cent  greater  than  the 
amounts  given  on  the  deflection  polygon  of  Fig.  39A.  The  maximum  deflection  due 
to  combined  shear  and  bending  effect  is  thus  0.93  +0.098  =  1 .028  inches,  for  the  point  7. 

The  approximate  formula  (39o)  gives  for  this  same  point 


A       5.37X12X56.52X122   . 
07  =        9X28X134000     '  + 

See  Art.  47  for  another  method  of  dealing  with  variable  moment  of  inertia,  by 
treating  the  quantities  MAx/I  as  elastic  loads  and  making  the  pole  distance  equal  to  E. 


CHAPTER   VIII 
DISPLACEMENT  INFLUENCE  LINES  FOR  STATICALLY  DETERMINATE  STRUCTURE! 

ART.  40.     INFLUENCE   LINES   FOR  ELASTIC   DISPLACEMENTS 

Deflection  influence  lines  could  not  receive  adequate  treatment  in  Chapter  T\ 
because  these  depended  on  a  knowledge  of  deflection  polygons.  The, latter  were  taker 
up  in  Chapters  VI  and  VII.  The  subject  is  now  treated  from  the  most  general  aspec 
covering  the  influence  line  for  any  elastic  displacement. 

Proceeding  from  Maxwell's  law,  Professor  Mohr,  in  1875,  first  developed  displacemen' 
influence  lines.  The  application  results  from  a  consideration  of  two  conventiona 
loadings,  of  unit  work  each,  applied  to  any  frame  so  that  one  of  the  loadings  represent; 
the  desired  influence  at  some  point  n,  and  the  other  loading  is  a  vertical  moving  loac 
P  =  l  applied  at  any  load  point  m  of  the  loaded  chord.  This  includes  all  cases  of  con- 
ventional loadings  given  in  Art.  9,  besides  the  simpler  problems  pertaining  to  displace 
ments  of  points. 

As  applied  to  any  simple  beam  or  truss,  the  following  theorem  is  now  established 
A  deflection  polygon,  drawn  for  a  load  unity  acting  in  a  fixed  direction  on  some  detiniti 
point  n  of  any  frame,  is  the  deflection  influence  line  for  the  deflection  of  the  point  n,  in  thi 
flxed  direction  for  any  system  of  parallel  moving  loads  applied  to  the  loaded  chord.  Wher 
the  system  of  moving  loads  does  not  consist  of  parallel  loads,  then,  of  course,  no  influence 
line  is  possible. 

To  prove  this  theorem,  the  simple  beam,  Fig.  40A,  is  loaded  with  a  vertical  loac 
Pn  =  l  applied  at  the  point  n.  A  deflection  polygon,  drawn  for  this  case  of  loading 
will  then  be  the  deflection  influence  line  for  the  point  n,  according  to  the  following 
demonstration.  The  method  of  drawing  the  deflection  polygon  is  given  in  Art.  39. 

The  moment  diagram  is  first  drawn  for  the  conventional  loading  Pn  =  l  kip,  appliec 
at  n  and  thus  making  the  maximum  ordinate  under  n  equal  to  Mn  =  l-aa'/l=4.8  kip 
ft.  The  symbol  M  is  used  to  indicate  a  moment  due  to  a  conventional  unit  load. 

This  diagram  is  divided  into  suitable  sections  (2  ft.  apart)  and  the  moment  areas 
MAx  are  computed  and  treated  as  elastic  loads  w,  for  which  the  deflection  poly- 
gon A "B"  is  finally  drawn,  all  as  shown  in  Fig.  40A. 

Assume  the  modulus  £'=28000  kips  per  sq.  inch,  7=2087,  in  inches,  M  in  kip 
feet,  and  Jo?  in  feet.  Then  the  pole  distance  for  the  force  polygon  becomes  H  =£7/144 
when  the  w  loads  are  made  equal  to  MJx,  by  Eq.  (39j),  for  deflections  of  actual  size, 
However,  the  scale  of  lengths  of  the  drawing  was  made  1 :36  and  the  deflections  should 
appear  200  times  actual,  hence  H  must  be  made  £7/200X36X144=56.35  w  units. 

122 


40 


DISPLACEMENT   INFLUENCE   LINES 


123 


With  a  pole  H=  56.35  and  the  w  loads  MAx,  construct  the  force  and  equilibrium 
polygons  as  shown,  and  the  deflection  polygon  so  found  represents  the  deflection  influence 
ine  for  the  point  n  according  to  the  following  application  of  Maxwell's  law. 

The  vertical  deflection  of  any  point  m,  for  the  given  load  Pn  =  l,  is  represented  by 
:he  vertical  ordinate  rjm  of  the  deflection  polygon  A"B",  vertically  under  the  point  m. 
For,  by  Maxwell's  law,  rjm=dmn=dnm,  or  in  words,  T?OT  equals  the  elastic  displacement 
:>f  the  point  n  for  a  unit  load  Pm  acting  at  the  point  m.  Hence  m  being  any  position 

|g.iKip. 
yn 


24-" 

SOIb.  I  Beam  ,  I  =2O87.z  in* 

;  I 

_  •»»»•  • 

% 

,                                                                               i-    fcw 

7 

rl 

M    DIAGRAM  FOR  P  =  I  KIP. 


B 


Scale  for  len^thsland  momenta. 


DEFJ.ECTIO|^  INFLL|ENCE  L|NE  FOR; POINT  n. 


B 


Deflections    zoo  times  actual. 

FIG.  40A. 

Df  the  moving  load  point,  it  follows  that  all  ordinates  rjm  represent  deflections  for  the 
point  n  due  to  a  unit  load  at  the  variable  point  m.  This  makes  the  polygon  A"B"  the 
iesired  deflection  influence  line  for  the  point  n. 

Therefore,  for  any  train  of  moving  loads,  the  deflection  at  the  point  n  becomes 

(40A) 


Since  Maxwell's  law  is  generally  applicable  to  any  case  of  conventional  loading, 
whether  for  a  single  point,  a  pair  of  points  or  a  pair  of  lines,  as  illustrated  in  Art.  9, 


124 


KINETIC  THEORY  OF  ENGINEERING  STRUCTURES      CHAP.  VIII 


therefore,  the  above  application  affords  a  solution  for  any  displacement  influence  line 
other  than  those  for  simple  vertical  deflections.  However,  the  train  of  loads  must 
consist  of  parallel  forces  P,  as  otherwise  no  influence  line  is  possible. 


ART.  41.     SPECIAL   APPLICATIONS   OF   DISPLACEMENT   INFLUENCE   LINES 

(a)  Required  the  deflection  influence  line  for  the  point  n  of  the  simple  cantilever 
beam,  Fig.  4lA.  The  deflection  polygon  is  drawn  according  to  the  method  of  Art.  31) 
and  illustrated  in  Art.  40,  for  the  conventional  loading  Pn  =  l  at  the  point  n.  This 
then  becomes  the  deflection  influence  line  for  the  point  n. 


FIG.  4lA. 

The  beam  is  anchored  at  A  and  supported  by  a  hinged  bearing  at  R  and  a  roller 
bearing  at  C.  The  point  n  is  the  hinged  connection  at  the  end  of  the  cantilever  arm 
Bn.  The  following  moments  M  and  reactions  R,  resulting  from  the  conventional 
loading  Pn  =  l,  are  then  found: 


A  =  --^,      B=- 

L\ 
MA=0,  MK-- 


^ 

^   —V 

Mn=0. 


The  M  diagram  is  constructed  by  making  the  ordinate  at  B'  equal  to  12,  and  after 
dividing  this  diagram  into  suitable  sections  and  computing  the  loads  w\  to  w8,  the 
deflection  polygon  is  easily  drawn,  observing  the  method  followed  in  Art.  40. 

The  closing  line  of  the  deflection  polygon  is  then  found  to  be  A"B"C",  which 
completes  the  influence  line.  The  point  B"  is  the  intersection  of  the  deflection  line 


ART.  41 


DISPLACEMENT   INFLUENCE  LINES 


with  the  vertical  through  B  and  n"C"  is  a  straight  line.     Upward  deflections  are  negative. 

The  support  at  A  is  elastic  and  its  displacement  oAa,  due  to  the  conventional 
loading  Pn=-l,  should  be  computed  from  Eq.  (4A)  including  temperature  effect  if 
desirable.  This  displacement  is  then  applied  at  the  point  A"  and  the  final  closing 
line  Ai"B"C\"  i-s  thus  made  to  include  this  effect.  By  applying  any  train  of  loads  the 
resulting  actual  deflection  of  the  point  n,  according  to  Eq.  (40A)  becomes  dn  =  HPi). 

For  any  case  of  variable  moment  of  inertia  of  the  given  beam  the  pole  distance  H 
is  made  variable,  as  was  done  in  Fig.  39A. 

(b)  Given  a  simple  truss,  Fig.  41  B,  on  two  supports,  to  find  the  displacement  influ- 
ence line  for  any  panel  point  n  for  displacements  dn  in  the  fixed  direction  rm7.  The 
loading  is  to  consist  of  a  system  of  vertical  concentrated  loads  P  acting  on  the  bottom 
chord. 


FIG.  41s. 


Apply  the  conventional  loading  Pn  =  l  in  the  given  direction  nn'  and  compute  the 
reactions  H,  A  and  B  for  this  load,  then  determine  t-he  stress  S\  and  resulting  change 
Jl  in  length,  for  each  member.  Finally  compute  the  w  loads  for  the  several  panel  points 
and  draw  the  deflection  polygon  for  the  loaded  chord  (here  the  bottom  chord)  using  any 
of  the  methods  of  Chapter  VI,  but  preferably  the  one  given  in  Art.  36  (c)  .  The  direction 
of  the  w  loads  is  always  taken  parallel  to  the  system  of  loads  P,  hence  the  displacement 
influence  ordinates  jj  will  be  vertical  ordinates,  measured  vertically  under  the  respective 
loads  P.  This  is  necessary  because  the  direction  of  the  deflections  is  determined  by 
the  direction  assigned  to  the  w  loads. 

The  proof  that  the  polygon  A  'B'  is  the  desired  influence  line  again  follows  from 
Maxwell's  law,  viz.,  for  any  ordinate  rjm  =dmn  =  dnm. 

The  displacement  dn  of  the  point  n  in  the  direction  nn',  caused  by  the  system  of 
vertical  loads  P  is  then 


(c)  Given  the  three-hinged  arch,  Fig.  41c,  to  find  the  influence  line  ior  the 
angular  rotation  3n  between  the  two  lines  An  and  Bn.  The  train  loads  are  to  be  carried 
bv  the  top  chord. 


126 


KINETIC  THEORY  OF  ENGINEERING  STRUCTURES      CHAP.  VIII 


The  conventional  loading  now  becomes  one  of  loading  a  pair  of  lines  with  a  positive 
moment  equal  to  unity  applied  to  each  line  in  the  direction  in  which  the  angle  J  would 
increase  for  vertical  loads  acting  on  the  top  chord. 

The  horizontal  thrust  for  this  conventional  loading  is  obtained  by  taking  moments 
about  the  crown  n  and  gives  H  =  (2-{-AV)/<2f  and  the  reactions  A  and  :i  are  found  by 
taking  the  moments  of  the  external  forces  about  B  and  A  respectively. 

Then  for  this  case  of  conventional  loading  determine  the  stresses  S\  and  changes 
Al  in  the  lengths  of  the  members  and  compute  the  w  loads  for  the  half  span,  which  is 
all  that  is  required  for  a  systematic  structure. 


FIG.  41c. 

The  moment  diagram,  drawn  for  the  w  loads,  will  then  be  the  displacement  influence 
line  for  the  change  dn  in  the  angle  J.  The  dn  for  any  particular  set  of  loads  is  then 
dn  =2Prj,  measured  in  arc. 

The  above  problems  will  suffice  to  show  the  perfectly  general  solution  of  displace- 
ment problems  afforded  by  the  application  of  Maxwell's  law.  These  problems  can  also 
be  solved  analytically  with  the  aid  of  Mohr's  work  equation  as  indicated  in  Art.  G. 


CHAPTER  IX 
INFLUENCE  LINES  FOR  STATICALLY  INDETERMINATE  STRUCTURES 

ART.  42.     INFLUENCE    LINES   FOR   ONE    REDUNDANT   CONDITION 

The  principle,  illustrated  in  Chapter  VIII,  for  drawing  deflection  influence  lines,  is 
easily  applied  to  the  construction  of  the  influence  line  for  any  external  or  internal 
redundant  condition. 

Eqs.  (7n)  and  (So)  for  one  redundant  condition  and  a  single  external  force  P  =  1 
become 

' 


a  a0  aa  f- 

and  •     •     <42i> 

"a  —     1  '  "ma  ~       XaOaa    =  Xaf)a 

after  substituting  the  value  da  as  obtained  from  Eq.   (4A)  and  neglecting  temperature 
and  abutment  displacements. 

This  gives  Xa  in  terms  of  displacements  or  stresses,  whichever  is  preferred,  as 


Y  - 

' 


where  the  redundant  may  be  external  or  internal. 

If  the  single  load  P  =  l  is  vertical  and  applied  to  some  panel  point  of  the  loaded 
chord  then  Eq.  (42s)  represents  the  influence  line  for  Xa  for  a  unit  load  applied  at  any 
panel  point  m. 

A  deflection  polygon,  drawn  for  the  conventional  loading  or  condition  Xa  =  l,  will 
give  the  ordinates  r}m=dma=dam  =  'ZS0Sap  for  the  displacement-  of  the  point  a,  for  a  unit 
load  at  any  point  m  of  the  loaded  chord.  Hence  the  deflection  polygon  drawn  for 
condition  Xa  =  1  is  the  Xa  influence  line  provided  all  the  ordinates  are  divided  by  the 
constant  denominator  daa+pa  =  '£Sc?o  +  pa,  The  constant  displacement  daa=2>Sa2o  is 
found  by  computation  or  otherwise,  and  pa  is  a  given  constant  depending  on  the  length 
and  section  of  the  redundant  member.  When  Xa  is  a  reaction  then  ,oa=0  for  the  case 
of  immovable  supports. 

Accordingly  the  following  theorem  may  be  stated:  The  ordinates  to  the  influence 
line  of  any  redundant  Xa  are  some  constant  factor  p.  —  1  ~  (d^  +pa)  times  the  ordinates  to 
a  deflection  polygon  drawn  for  the  loaded  chord,  for  the  conventional  loading  Xa  =  l  (P=0), 
applied  to  the  principal  system. 

127 


128 


KINETIC  THEORY  OF  ENGINEERING  STRUCTURES 


CHAP.  IX 


(a)  When  the  redundant  X<j  is  an  internal  condition,  as  illustrated  in  Fig.  42A,  the 
above  theorem  is  applied  as  follows:  If  the  direction  of  stress  in  the  redundant  can  be 
anticipated,  then  this  should  be  done,  otherwise  an  arbitrary  assumption  must  be  made 
and,  if  it  was  erroneous,  the  resulting  value  of  the  redundant  Xa  will  turn  out  negative 
with  respect  to  S0.  In  any  case  the  conventional  loading  Xa  =  l,  being  external  work, 
should  be  so  applied  as  to  produce  positive  work  when  the  redundant  member  is  removed. 
In  the  illustration,  the  member  Xa  will  be  assumed  in  tension  so  that  J/a  is  a  positive 
elongation.  Hence,  the  unit  forces  Xa  =  1  must  be  applied  so  as  to  move  the  points  n 
and  HI  apart.  The  principal  system  is  the  entire  frame  exclusive  of  the  member  Xa. 

The  stresses  Sa  are  computed  or  found  from  a  Maxwell  diagram  and  from  these  the 
w  loads  and  deflection  polygon  for  the  loaded  chord  (which  was  taken  to  be  the  bottom 
chord  in  the  illustration)  are  determined  by  one  of  the  methods  in  Chapter  VII. 


The  deflection  polygon  so  obtained  may  be  represented  by  the  broken  line  A'n'rii'B', 
with  ordinates  i)m=dma=dam,  which,  according  to  Eq.  (42B)  are  (daa+pa)  times  the  actual 


influence  line  ordinates  for  the  redundant  stress  Xa. 
system  of  moving  loads,  becomes 


Hence    the    stress  Xa,  for  any 


(42c) 


(b)  When  the  redundant  Xa  is  an  external  condition,  the  quantity  pa  becomes  zero 
for  the  assumption  da=0  for  immovable  supports,  and  this  gives  rise  to  a  slight  simpli- 
fication in  the  solution  of  problems  of  this  class. 

Eq.  (42n)  then  becomes  in  general 


Oaa 


(42D) 


The  illustration  of  a  simple  beam  on  three  supports  is  given,  in  Fig.  42B,  as  a  case 
of  one  external  redundant  condition  Xc.  Any  one  of  the  three  supports  might  be 
treated  as  the  redundant  one,  but  the  middle  support  C  is  here  chosen. 

The  conventional  loading  Xc  =  1  is  again  chosen  so  as  to  produce  a  positive  quantity 
of  applied  work.  The  deflection  polygon  A"C"B"  of  the  simple  determinate  beam  AB, 


ART.  42       INFLUENCE   LINES   FOR  INDETERMINATE  STRUCTURES 


125) 


which  now  becomes  the  principal  system,  is  found  for  this  load  Xc  =  l  exactly  as  was 
done  in  Art.  40,  Fig.  40A.  The  unit  load  is  applied  at  C  in  a  downward  direction 
because  the  deflection  of  this  point  of  the  principal  system  is  also  downward  when  the 
support  is  removed. 

Any  ordinate  rjm  of  this  deflection  polygon  is  equal  to  the  deflection  dmc=3cm,  which 
is  §cc  times  the  Xc  influence  line  ordinate  according  to  Eq.  (42D).  Also  the  special 
ordinate  f)c=dcc  is  given  by  the  same  deflection  polygon  and  is  a  constant  quantity. 
Hence  the  factor  /*  for  the  influence  line  Xc  is  equal  to  l/dcc  and  the  ratio  dmc/dcc 
determines  the  value  of  the  influence  line  ordinate  for  any  point  m. 


k 

im 


FIG.  42s. 

It  is  thus  seen  that  for  the  case  of  a  single  external  redundant  Xc  the  magnitude  of 
the  moment  MM=lilz/l  and  the  pole  distance  H  of  the  deflection  polygon,  do  not  affect 
the  influence  line  ordinates  Tim=dme/8ee,  though  they  do  affect  each  of  the  ordinates  $mc 
and  Scc  separately. 

For  any  system  of  moving  loads  the  required  redundant  reaction  becomes 


(42E) 


130  KINETIC  THEORY  OF  ENGINEERING   STRUCTURES          CHAP.  IX 


ART.  43.     INFLUENCE    LINES   FOR   TWO   REDUNDANT   CONDITIONS 

When  there  are  two  redundants,  then  two  deflection  polygons  may  be  drawn  for 
the  loaded  chord  of  the  principal  system  ;  one  for  condition  Xa  —  1  and  one  for  condition 
X\)-=\.  These  polygons  are  then  combined  into  an  influence  area  for  Xa  the  ordinates 
to  which,  times  a  certain  factor  /*„,  will  give  the  influence  line  ordinates  for  the  Xa 
influence  line.  Another  combination  similarly  made  for  Xb  will  give  ordinates  which 
when  multiplied  by  a  factor  fib,  will  give  the  ordinates  for  the'Afc  influence  line.  The 
analytic  solution  for  these  influence  lines  is  derived  from  Eqs.  (SD). 

Thus  for  two  external  redundants  with  da  and  05  both  zero  and  neglecting  temper- 
ature and  reaction  displacements,  Eqs.  (80)  give,  for  a  single  load  P, 


=  XaOaa 

........     .     .     (43A) 


These  equations  solved  for  Xa  and  A^  and  observing  that  ^=^06  by  Maxwell's 
law,  give 


__ 
°ma 


*  -p 

_  _  °bb  p  _i}mar  -p 

a—  —*  —  «    •  —  {ta^ma^ 

*  *    Oab  T)a 

°aa       °o6"F~ 
066 


- 

°mb      °ma  * 


wherein  dma  and  dmb  are  the  variable  ordinates  to  the  two  deflection  polygons  Xa  =  1 
and  Afe  =  l,  respectively,  for  any  panel  point  m.  The  other  quantities  are  all  constants 
given  by  the  same  deflection  polygons. 

These  equations  for  a  load  P  —  l,  are  the  equations  for  the  influence  lines  Xa  and 
Xb.  Their  numerators  then  give  variable  ordinates  which  may  be  called  r)ma  and  i)mb 
while  the  constant  denominators  ija  and  yb  may  be  used  as  influence  factors  fta  =  l/T)a 
and  fjtb  =  l/r)b  for  the  two  influence  lines  Xa  and  Xb  respectively. 

For  any  train  of  moving  loads  the  redundant  reactions  are  expressed  by 

ma     } 

\  ...........     (43c) 

inb     • 

The  problem  of  a  simple  beam,  Fig.  43A,  continuous  over  four  supports,  is  chosen 
to  illustrate  this  case. 

The  redundant  supports  are  considered  removed  and  the  principal  system  then 
becomes  a  simple  beam  on  two  supports.  In  the  present  case  the  supports  Xa  and  A"0 
are  treated  as  redundants,  although  any  two  of  the  four  reactions  could  be  thus  regarded. 


ART.  43         INFLUENCE  LINES  FOR  INDETERMINATE  STRUCTURES 


131 


The  moment  diagrams  for  the  two  conventional  loadings  Xa  =  l  and  Xh=--l  are  now 
drawn  and  from  these  the  deflection  polygons  are  constructed  precisely  as  illustrated 
in  Fig.  40A.  The  two  poles  H  are  made  equal  and  as  a  check  on  the  drawing  <?a6'must 
equal  o&a.  When  the  structure  is  a  frame,  then  the  deflection  polygons  for  the  loaded 
chord  are  drawn  as  described  in  Art.  37. 

The  deflection   polygons  so  obtained  are  the  influence  lines  for  the  deflections  dn 


FIG.  43A. 


and  #6  of  the  two  redundant  supports.     For  immovable  supports  these  are  zero  and 
Eqs.  (43A)  apply  to  any  case  of  loading. 

To  obtain  the  influence  areas  for  the  redundants,  the  two  deflection  polygons  are 
combined  into  two  influence  areas  by  computing  the  ordinates  from  Eqs.  (43s)  as  follows: 


b^ 
Obb 


132  KINETIC  THEORY   OF   ENGINEERING  STRUCTURES         CHAP.  IX 

The  factors  /*«,  and  fj.b  are  the  reciprocals  of  the  denominators  of  Eqs.  (43B)  all  the 
values  for  which  are  given  on  the  two  deflection  polygons.  The  ordinates  rjma  give, 
when  plotted,  the  shaded  area  C'D',  which  is  the  influence  area  for  Xa.  The  ordinate 
T)a,  under  the  support  Xa,  gives  the  influence  factor  //a  =  l/V  In  similar  manner  the 
ordinates  rjmb  furnish  the  shaded  area  C"D",  which  is  the  influence  area  for  Xb  with 
the  factor  /j.b  =  I/fa,  obtained  from  the  ordinate  rj&  under  the  support  Xb. 

A  purely  graphic  construction  of  the  influence  area  Xa  is  given  by  Mueller-Breslau  as 
follows:  Draw  the  da  deflection  polygon  for  the  wa  loads,  using  any  convenient  pole  H. 
Then  for  the  w^  loads  draw  an  equilibrium  polygon  passing  through  the  three  points 
C',  B'  and  D' '.  This  is  perhaps  the  best  and  most  convenient  solution  and  is  also 
illustrated  in  Fig.  43A. 

The  da  and  db  deflection  polygons  are  first  drawn  as  above  described  with  any 
assumed  pole  distance  which  may  be  the  same  for  each  polygon.  Then  draw  the  db 
line  through  the  three  points  C',  B'  and  D',  and  also  draw  the  da  line  through  the  three 
points  C",  A"  and  D".  The  shaded  areas  thus  formed  represent  the  Xa  and  Xb  influence 
areas. 

To  find  the  new  poles  0'  in  the  two  force  polygons,  which  are  necessary  in  drawing 
an  equilibrium  polygon  through  three  given  points,  proceed  as  follows:  In  the  db  force 
polygon,  draw  OK\\C"D"  and  OKf^C^W7  thus  determining  the  points  K  and  K'. 
Then  draw  KO'\\C'D'  and  K'0'\\C'B'  and  the  new  point  0',  thus  found  by  the  inter- 
section of  K'O'  and  KO' ',  will  be  the  required  pole  for  drawing  the  db  line  through  C', 
B'  and  D'. 

The  pole  0'  in  the  da  force  polygon  is  found  in  a  similar  way  as  indicated  in  the 
figure,  using  the  points  F  and  F'.  The  new  pole  0' ' ,  found  by  the  intersection  of  FO' 
and  F'O',  serves  to  draw  the  da  line  through  the  points  C",  A"  and  D". 

This  construction  is  based  on  the  fact  that  the  influence  of  a  load  at  B  on  Xa  must 
be  zero,  likewise  for  a  load  at  A  on  Xb. 

The  above  methods  are  not  practicable  for  more  than  two  redundants,  and  the 
following  method  of  simplification  is  given  for  multiple  redundancy. 


ART.  44.     SIMPLIFICATION   OF   INFLUENCE   LINES  FOR  MULTIPLE  REDUNDANCY 

In  many  problems  of  this  class  it  is  possible  to  so  choose  the  redundant  forces 
X  that  they  will  have  a  common  point  of  application.  Then  for  certain  directions 
of  the  X'8  the  d  coefficients  bearing  different  subscripts  and  appearing  as  factors  of 
the  X's  in  Eqs.  (So)  may  be  reduced  to  zero.  Whenever  this  is  possible  then 
vao-—0ba=--0,  dac=dca-—0,  ^bc=^^d>=0',  etc. ;  and  the  following  simple  equations  are 
obtained1. 


9*  X^  7~>       S»  ~V    &  \^  Z?     Af*     1    J}1 

Ob  =  Ztr'mOmb  —  AfcOfcfc  —  Lttibar  +  0bt 


(44A) 


ART.  44        INFLUENCE  LINES  FOR  INDETERMINATE  STRUCTURES  133 

involving  only  one  redundant  X  in  each  equation.  The  same  assumptions  may  be 
applied  to  Eqs.  (7n). 

This  method  of  analysis  was  first  introduced  by  Professor  Mueller-Breslau,  and 
serves  a  most  valuable  purpose,  especially  when  applied  to  fixed  arches. 

As  will  be  shown  presently,  this  disposition  of  the  redundants  is  easily  made 
when  their  number  does  not  exceed  three.  Beyond  this  number  the  analysis  leads 
to  many  complications.  Fortunately,  the  important  cases  of  redundancy  are  nearly 
always  limited  to  from  one  to  three  redundants,  and  then  there  is  no  difficulty  in 
following  the  scheme  here  proposed. 

The  solution  of  Eqs.  (So)  for  simultaneous  values  of  the  X's  is  thus  avoided  and 
the  other  chances  for  serious  errors  are  greatly  lessened. 

It  is  usually  expedient  to  treat  temperature  stresses  and  abutment  displacements 
separately  from  the  primary  stresses,  and  then  Eqs.  (44A)  become  for  load  effects  only 
and  immovable  abutments, 

Z2->im°ma           -\r         ^*m0mb  /it     \ 

a= — ^ — ;    xb  = — 5 — ;    etc (44s) 


*«!=!--;         *6i=l-rJ     etc (44c) 

Oaa  Obb 

By  placing  2.RaJr=0,  da  also  becomes  zero,  likewise  for  S-fi^Jr  and  db,  a  circum- 
stance which  should  be  noted  in  writing  Eqs.  (44s)  and  Eqs.  (44c). 

In  all  of  the  following  investigations,  the  X's  will  represent  either  a  single  force  or 
a  couple  applied  to  a  principal  system.  The  d's  in  the  first  case  will  then  represent 
linear  displacements  and  in  the  second  case  they  will  be  angular  distortions  expressed 
in  arc.  Thus,  if  a  redundant  Xa  is  applied  to  the  point  a  then  ^  is  the  displacement 
of  the  point  a  in  the  direction  of  Xa  for  a  force  Xa  =  1.  When  Xa  is  a  moment  or  couple, 
then  daa  becomes  the  angular  rotation  of  a  rigid  principal  system  as  produced  by  a 
couple  Xa--=l.  The  displacements  da,  db,  etc.,  and  the  conventional  loadings  Xa  =  l', 
Xb  =  l,  etc.,  are  always  positive  in  the  opposite  sense  in  which  the  redundants  Xa,  Xb, 
•etc.,  are  supposed  to  act. 

It  should  also  be  noted  that  in  all  cases  here  considered,  the  points  a  and  b  are 
coincident  and  these  designations  are  retained  merely  to  distinguish  the  particular  forces 
from  each  other. 

The  Eqs.  (44A)  to  (44c)  are  equally  applicable  to  graphic  and  analytic  solutions, 
but  the  latter  method  is  useful  only  when  there  are  not  more  than  three  redundant 
conditions,  and  great  accuracy  is  desired.  The  graphic  method  is  less  laborious  and 
leads  to  a  more  comprehensive  presentation. 

The  above  will  now  be  applied  to  general  cases  of  two  and  three  redundants. 

(a)  Structures  having  two  redundant  conditions.  Here  the  two  redundant  forces 
can  always  be  applied  at  the  same  point  and  the  case  can  be  solved  by  applying  Eqs. 
(44s)  and  (44c),  provided  the  directions  of  Xa  and  Xb  are  so  chosen  that  #&,  =£«&=(). 


134 


KINETIC    THEORY  OF  ENGINEERING  STRUCTURES 


CHAP.  IX 


To  accomplish  this,  assume  the  direction  of  Xa  as  may  seem  most  convenient,  and 
find  the  displacement  a\a  of  the  point  a  for  a  load  Xa  =  l,  Fig.  44A.  Now  take  Xb 
acting  at  the  same  point  a  and  at  right  angles  with  aid. 

Then  ^=0,  because  it  represents  the  displacement  of  the  point  6  in  the  direction 
of  Xj,  when  only  the  force  Xa  =  l  is  active.  Therefore,  dab  also  becomes  zero 
by  Maxwell's  law.  If  then  the  displacement  6j6  of  Xb  is  found  for  the  load  Xb  =  l, 
this  in  turn  must  be  perpendicular  to  Xa  because  dab  is  zero.  This  always  fur- 
nishes a  valuable  check  on  the  solu- 
tion when  the  graphic  method  is 
employed. 

The  example  of  a  two-hinged 
arch  with  a  column  support  at  the 
center,  Figs.  44B  to  44D,  is  used  to 
illustrate  the  application  of  the 
method  to  two  redundants. 

Fig.  44B  shows  the  given  struc- 
ture with  hinged  supports  at  A,  B 
and  C,  thus  involving  two  external 
redundants.  By  removing  the  sup- 
port at  A  the  structure  becomes 
determinate,  involving  a  simple  truss 
on  supports  B  and  C,  and  an  over- 
hanging cantilever  AC'.  The  redun- 


FIG.  44A. 


dant  forces  Xa  and  Xb  are  then  applied  to  the  principal  system  at  A  and  are  treated 
as  external  forces. 

The  first  redundant  Xa  is  assumed  to  act  horizontally  and  the  second  redundant  Xb 
is  applied  at  the  same  point  and  making  the  angle  6  with  Xa,  such  that  dab=dba=0. 

Fig.  44c  represents  the  conventional  loading  Xa  =  l  and  a  Williot-Mohr  displace- 
ment diagram,  drawn  for  this  loading,  will  furnish  the  displacement  aat  for  the 
point  A  and  the  displacement  mma  for  the  point  m.  Hence  dao  becomes  the  pro- 
jection of  act!  on  the  direction  of  Xa  and  <?„*,  is  the  projection  of  mma  on  the  direction 
of  the  force  Pm. 


ART.  44        INFLUENCE  LINES  FOR  INDETERMINATE  STRUCTURES  135 

Similarly  in  Fig.  44D,  another  displacement  diagram  _drawn  for  the  conventional 
loading  Xb  =  l  and  acting  at  90°  with  the  displacement  aal,  gives  the  displacements 
661  and  mmb  from  which  the  values  du,  and  dmb  are  found  by  projecting  the  displace- 
ments on  the  directions  of  their  respective  forces  Xb  and  Pm. 

The  angle  6  is  thus  graphically  determined,  and  as  a  check,  the  displacement  66t 
must  be  perpendicular  to  Xa. 

Thus  having  the  constants  dm  and  dbb  and  the  displacements  dma  and  dmb  for  any 
pin  point  m,  the  influence  produced  by  any  load  Pm,  according  to  Eqs.  (44s)  becomes 

Zi  mUma  j          v        PmOmb  /A4     -. 

a=  and      Xb=—  ^  —  ,  ........     (44D) 

daa  Obb 

where  dma  is  negative  with  respect  to  the  force  Pm,  as  may  be  seen  from  Fig.  44c. 
The  values  Xa  and  Xb  for  any  set  of  loads  become 

Z^imflma    ,  _j     v"        ^^mPmb  /AA    \ 

a=  —  5  -  and  Xb  =  —  r  -  .........     (44E) 


and  the  several  values  dma  and  dmh  are  found  from  the  two  displacement  diagrams 
above  described. 

When  the  load  P  =  1  is  vertical  then  the  values  in  Eqs.  (44D)  represent  influence 
line  ordinates  for  any  point  m  for  the  redundants  Xa  and  Xb.  The  same  displacement 
diagrams  will  furnish  all  values  dma  and  dmb  for  finding  all  the  influence  line  ordinates 
for  both  redundants. 

For  a  uniform  rise  in  temperature  of  t°,  the  point  A  will  move  horizontally  an 
amount  etl  and  the  projections  of  this  displacement  on  the  directions  of  Xa  and  Xb 
will  then  be  respectively  dat  =  etl  and  dbt  =  etl  cos  0,  whence 


1       £M  1  V  1       &    COS 

^l-r-       and      Xbt  =  l  — 


(b)  Structures  having  three  redundant  conditions.  The  most  prevalent  case  of 
three  redundants  is  a  fixed  arch,  and  hence  this  style  of  structure  is  chosen,  as  an 
illustration,  see  Figs.  44E  to  44j. 

Fig.  44E  represents  any  general  arch  with  fixed  supports.  It  is  transformed  into 
a  determinate  structure  by  removing  one  fixed  support  and  converting  the  frame 
into  a  cantilever  arm  to  be  treated  as  the  principal  system. 

If,  now,  this  support  is  replaced  by  three  redundant  conditions,  a  moment  l-Xa 
and  forces  Xb  and  Xc,  all  acting  on  the  rigid  disc  AJBO,  Fig.  44r,  the  stresses  in  the 
structure  will  remain  exactly  the  same  as  in  the  original  fixed_condition. 

Then  da  will  be  the  angular  rotation  of  the  rigid  disc  ABO  about  some  pole  0;  db 
will  be  the  linear  displacement  of  the  point  0  in  the  direction  of  Xb;  and  oc  will  be  that 


136 


KINETIC  THEORY   OF  ENGINEERING   STRUCTURES 


CHAP.  IX 


displacement  in  the  direction  of  Xc.     These  displacements  are  supposed  to  be  so  chosen 
that  the  moment  Xa  and  the  forces  Xb  and  Xc,  all   applied  to  the  rigid  disc  ABO,  will 

exert  precisely  the  same  effect  on  the  principal 
system  as  was  previously  produced  by  the  re- 
dundancy. 

The  pole  0  is  taken  as  the  instantaneous  center 
of  rotation  of  the  disc,  when  acted  upon  by  a 
moment  Xa  =  l.  If  this  pole  is  chosen  as  the  point 
of  application  for  the  redundants  Xb  and  Xc,  then 
dba=Q  and  £ca=0.  From  this  it  follows  that  the 
angular  displacements  <5a&  and  3ac  produced  by  the 
loadings  Xb  =  l  and  Xc  =  l  acting  on  the  disc,  must 
likewise  be  zero  if  the  drawing  or  computation  is 
correct.  Finally,  if  Xb  is  arbitrarily  chosen  in  a 
vertical  direction  and  Xc  is  taken  perpendicular  to 
the  displacement  which  the  point  b  (pole  0)  under- 
goes as  a  result  of  the  loading  .X&  =  1,  then 


PRINCIPAL    SYSTEM. 


CONDITION  X»l 


(J) 

FlGS.  44E,  F,  G,  H,  J. 


Hence  this  disposition  of  the  redundant  con- 
ditions will  cause  all  the  quantities  d  bearing  double 
subscripts  of  different  letters,  to  become  zero,  and 
the  simplified  Eqs.  (44A)  will  now  apply. 

When  the  abutments  are  considered  immov- 
able, da,  db,  and  dc  become  zero  and  Eqs.  (44B)  and 
(44c)  will  suffice  to  solve  the  fixed  arch.  For 
vertical  loads  these  same  equations  furnish  the 
ordinates  for  the  Xa,  Xb  and  Xc  influence  lines 
by  inserting  a  single  load  Pm  =  1 . 

In  proceeding  to  a  solution  it  is  best  to  apply 
a  unit  moment  e-l/e,  Fig.  44c,  to  the  points  A  and 
B,  representing  the  conventional  loading  Xa  =  l 
acting  on  the  principal  system.  A  Williot-Mohr 
displacement  diagram,  drawn  for  this  condition,  will 
furnish  the  displacements  A^A  and  BtB  of  the  two 
points  A  and  B.  The  pole  0,  being  the  instan- 
taneous center  of  rotation  of  the  rigid  disc,  is 
found  by  the  intersection  of  two  lines 


and  OB-LBiB.  Then  Xb  is  applied  at  0  in  a  verti- 
cal direction  and  Xc  is  made  active  at  0  and  in  a  direction  perpendicular  to  the 
displacement  found  for  0  from  the  displacement  diagram  drawn  ior  the  condition 
Xb=*l.  See  Figs.  44n  and  44j  for  the  conventional  loadings  Xj,  =  l  and  Xe  =  l. 


ART.  45        INFLUENCE  LINES  FOR  INDETERMINATE  STRUCTURES  137 


ART.  45.     STRESS   INFLUENCE    LINES   FOR   STRUCTURES   INVOLVING 

REDUNDANCY 

(a)  Multiple  redundancy.  The  previous  articles  42  to  44  show  how  influence  lines 
for  redundant  conditions  may  be  determined.  In  the  present  article  it  is  proposed  to 
present  the  methods  of  drawing  stress  influence  lines  for  the  members  of  any  struc- 
ture involving  redundant  conditions. 

Th?  general  equation  for  the  stress  in  any  member  of  the  principal  system  of  a 
statically  indeterminate  structure  is  Eq.  (7  A).  This  equation,  if  written  for  a  single 
external  load  Pm  =  l,  will  represent^  the  stress  influence  line  ordinate,  under  the  point 
ra,  for  the  member  S.  Hence,  if  £0  represents  the  stress  in  the  member  for  condition 
X  =0  and  Pm  =  1,  then  Eq.  (7  A)  gives  the  desired  influence  ordinate  for  any  point  ra  as 

i)m=S=S0-SaXa-SbXb-SeXe,etc.,     ......     (45A) 

where  rjm  is  the  algebraic  sum  of  the  partial  influences  S0,  Xa,  Xb,  Xc,  etc.,  on  the  stress 
of  a  certain  member  S,  due  to  a  load  Pm  =  1  acting  at  the  panel  point  ra.  In  other 
words,  a  separate  influence  line  may  be  drawn  for  each  term  of  Eq.  (4oA)  and  the 
algebraic  sum  of  the  several  influence  ordinates  under  a  certain  load  point  m  becomes 
the  stress  influence  ordinate  ym  for  that  particular  load  point.  Such  an  influence  line 
may  be  regarded  as  a  summation  influence  line  of  partial  stress  effects  instead  of  a 
summation  load  effect  as  originally  implied  by  the  definition  in  Art.  17. 

Accordingly,  for  any  system  of  external  parallel  loads,  the  total  stress  in  any 
member  is  represented  by 

.......     (45u) 


As  may  be  seen  from  Eq.  (45A)  such  a  stress  influence  line  will  always  necessitate 
drawing  as  many  influence  lines  as  there  are  redundants,  plus  one  for  the  S0  stress. 
However,  the  influence  lines  for  the  redundants  remain  unchanged  for  all  members  of 
the  same  structure  and  hence  need  be  drawn  only  once.  Also  the  stressesjSa,  Sb  and 
Sc  are  constants  for  a  given  member  but  vary  for  different  members.  The  S0  influence 
line  is  the  same  as  the  stress  influence  line  for  any  member  of  a  determinate  frame  and 
will  be  different  for  each  member.  See  Chapter  IV. 

Therefore,  it  is  advisable  to  draw  the  influence  lines  for  the  redundants  without 
the  Sa,  Sb,  Sc,  etc.,  factors  and  then  draw  the  ~So  stress  influence  line  for  any  particular 
member  S.  Finally  insert  for  the  X's  the  values  for  any  set  of  ordinates,  as  for  point 
k,  and  these  combined  with  the  stresses  Sa,  Sb,  Sc  in  Eq.  (4oA)  give  the  required  ordinate 
rjk  for  the  stress  influence  line  S.  The  several  ordinates  rj  for  all  panel  points  furnish 
the  required  influence  line.  However,  much  of  this  work  is  done  graphically  as  illus- 
trated in  Fig.  45A. 

The  example  is  a  two-hinged  framed  arch  with  a  column  supporting  the  crown,  thus 
involving  two  external  redundant  conditions  Xa  and  Xb  as  indicated  in  Fig.  45A.  The 
loads  are  vertical  and  applied  to  the  top  chord.  Required  to  find  the  stress  influence 
line  for  the  chord  ik. 


138 


KINETIC   THEORY   OF   ENGINEERING  STRUCTURES 


CHAP.  IX 


The  influence  lines  for  Xa  and  Xb  are  drawn  supposedly  by  the  method  given  in 
Art.  43.  The  ordinary  stress  influence  line  for  the  chord  ik  is  then  drawn  as  indicated 
and  called  the  S0  line,  meaning  that  it  is  the  influence  line  for  condition  X  =  0  with 
P  =  l.  This  line  A'i'k'B'  incloses  a  negative  area  representing  compressive  stress  in 
the  top  chord  of  a  simple  truss  AB. 


A 


B 


FIG.  45A. 


The  influence  lines  Xa  and  Xb  are  both  positive  and  the  stresses  Sa  and  Sb,  for 
the  member  ik,  are  found  to  be  negative.  The  ordinate  TJ^  for  the  panel  point  k  is 
found  from  Eq.  (45A)  as 

....".     T    .     .   ,.       (45c) 


giving  a  negative  residual  for  yk. 

In  like  manner  all  ordinates  for  the  S  influence  line  may  be  found  and  plotted  to 
give  the  shaded  area  A  'B'  ,  which  is  the  influence  area  for  the  chord  ik. 

The  multiplications  Sar)a,  etc.,  can  be  performed  graphically  by  laying  off  angles 
corresponding  to  tana=Sa  and  tan  ft=Sb  as  indicated  in  the  figure.  The  products 
Sarja  and  Sbyb  are  then  taken  off  directly  for  any  ordinate,  added  or  subtracted  graphically 
and  the  sum  Saya+SbT)b  is  then  applied  to  the  ordinate  S0  in  the  proper  direction.  The 
angles  a  and  /?  are  constant  for  the  same  member  but  vary  for  different  members. 


ART.  45        INFLUENCE  LINES  FOR  INDETERMINATE  STRUCTURES 


139 


The  stresses  Sa  =tan  a  and  Sb  =tan  /?  are  found  from  Maxwell  diagrams  or  by  computation. 

It  is  thus  seen  that  all  loads  between  A  and  C  produce  compression  in  the  member 
ik,  while  all  loads  between  C  and  B  will  produce  tension. 

The  stress  resulting  from  any  system  of  loads  is  then 


etc.=2Pi?, 


(45D) 


observing  the  sign  of  rj  for  each  particular  point. 

Since  the  large  majority  of  practical  problems  of  redundancy  do  not,  and  should 
not,  involve  more  than  one  redundant  condition,  the  general  case  is  here  treated  in 
less  comprehensive  manner.  For  a  special  method  of  deriving  the  influence  line  for  a 
web  member  from  those  of  two  adjacent  chords,  see  Art.  52. 

(b)  One  redundant  condition  according  to  Eq.  (45A)  gives  rise  to  the  simple  stress 
equation 

/"cT  \ 

(45E) 


*jm       *^       ^o       *^&^*-a       *^a  I   ct          -^-ct 
V>a 

which  makes  it  possible  to  represent  the  S  influence  area  as  the  area  inclosed  between 
the  Xa  influence  line  and  the  S0/Sa  influence  line,  which  latter  is  drawn  with  as  much 
ease  as  the  simple  S0  line.  The  resulting  S 
influence  area  will  then  require  a  factor  Sa 
applied  to  all  its  ordinates. 

These  influence  areas  may  be  drawn  in 
either  of  two  ways:  First,  by  plotting  both 
the  Xa  and  the  S0/Sa  lines  from  a  common 
straight  base,  observing  the  rule  that  areas 
above  this  base  are  negative,  see  Fig.  45u. 
The  S  line  ordinates  TJ  will  then_  be  measured 
between  the  Xa  line  and  the  S0/Sa  line  and 
these  ordinates  will  be  positive  when  measured 
down  from  the  Xa  line.  Second,  by  first 
plotting  the  Xa  line  from  a  straight  base  and 

then  constructing  the  S  area  by  applying  the  +S0/Sa  ordinates  y'  down  from  the  Xa 
line,  see  Fig.  45c.  In  this  case  the  +Xa  ordinates  are  best  plotted  above  the  base  so 
that  the  4-  TJ  ordinates  of  the  S  area  will  appear  below  the  base  in  the  customary  way. 

The  first  method  is  more  generally  used,  though  the  second  leads  to  a  very  interesting 
property  by  which  the  S  line  may  be  derived  from  the  Xa  line  when  one  point  of  the  S 
line  and  its  zero  points  are  known. 

This  property  is  shown  in  Fig.  45c,  where  it  is  seen  that  over  any  distance,  as  m'B', 
for  which  the  ~S~0/Sa  line  is  straight,  the  corresponding  elements  of  the  S  line  and  the 
Xa  line  will  intersect  in  points  b',  b" ,  &'"_which  are  in  a  straight  line  parallel  to  P  and 
passing  through  the  end  zero  point  of  the  S0/Sa  line. 

Further  details  and  applications  are  not  given  here,  as  these  will  be  illustrated  in 
connection  with  specific  problems  in  Chapter  X. 


CHAPTER  X 

SPECIAL  APPLICATIONS  OF  INFLUENCE  LINES  TO  STATICALLY  INDETERMINATE 

STRUCTURES 

ART.  46.     SIMPLE    BEAM   WITH    ONE    END    FIXED    AND    OTHER   END  SUPPORTED 

The  solution  by  influence  lines  is  illustrated  by  Figs.  46A,  using  the  general  method 
developed  in  Art.  45B. 

Applying  the  criterion  of  Eq.  (3c)  to  this  structure,  where  m  =  l,  2r=4  and  2p=4, 
gives  ra  +  2r-2p  =  l.  Hence  there  is  one  redundant  condition  and  by  Eq.  (SB)  this 
condition  must  be  external. 

Accordingly,  the  reaction  Xa,  acting  at  the  expansion  end  B,  is  chosen  as  the 
redundant  support,  reducing  the  beam  to  a  simple  cantilever  arm  as  the  principal 
determinate  system. 

The  problem  is  considered  solved  when  the  shear  and  moment  influence  lines  are 
found  for  any  point  n  of  the  beam.  The  case  of  direct  loading  is  assumed. 

The  Xa  influence  line  is  represented  by  Eq.  (42o)  as 

i   ^ 

„         1  .  Ont,  ,  . 


where  dma  is  the  variable  influence  line  ordinate  at  any  point  m  with   [1  =  1/8^  is  the 
constant  factor,  when  dma  and  daa  may  be  acutal  or  measured  to  the  same  scale. 

The  equations  for  moment  and  shear  at  any  point  n  and  for  the  vertical  reaction 
A  ,  are  given  by  Eqs.  (?A)  written  in  the  form  of  Eq.  (45E)  as 


—  ~J          A    Y    --  4  Y  \ 

— **•        Aa-A-a  ~  -^rtl  ~r~       -A-a 
\Aa 


(46B) 


wherein  M0  is  the  moment  about  the  point  n  produced  in  the  principal  system  by  a 
load  unity  acting  at  any  point  m  of  the  structure.  Ma  is  the  moment  at  the  point  n 
due  to  the  conventional  loading  Xa  =  l.  Similar  definitions  follow  for  Q~0,  Q*,  also  A^0 
and  Aa. 


140 


ART.  46  SPECIAL  APPLICATIONS  OF  INFLUENCE  LINES  141 

The  following  solution  by  influence  lines  is  based  on  these  Eqs.  (46B),  when  the 
influence  line  for  Xa  is  known  or  found  from  Eq.  (46A) . 

The  ordinates  dma  are  ordinates  of  a  deflection  polygon  drawn  for  the  conventional 
loading  Xa  =  l  kip,  being  a  load  unity  acting  downward  at  the  point  B  when  the  beam 
is  treated  as  a  cantilever.  This  deflection  polygon  is  drawn  as  described  in  Art.  40, 
and  as  shown  in  Fig.  46Aa. 

The  M  moment  diagram  for  the  conventional  loading  Xa  =  l,  is  drawn  by  making 
the  end  ordinate  A^C'  =1  and  drawing  C^B'.  This  end  ordinate  may  be  measured  to 
any  convenient  scale  and  was  drawn  to  half  the  scale  of  lengths,  making  all  the  moment 
ordinates  to  half  scale  of  lengths.  The  figured  ordinates  represent  moments,  in  kip  feet, 
and  are  scaled  at  points  Jz=2  ft.  apart. 

The  areas  w  =Jfdx  are  now  computed  and  treated  as  elastic  loads  for  the  construc- 
tion of  the  force  polygon  (a)  and  the  deflection  polygon  (b)  when  the  pole  distance  H=EI 
Takino-  E  =28000  kips  and  7=2087  in.4  and  reducing  Ax  to  inches  and  M  to  kip 
inches    then  the  pole  would  be  ™™£™.      When  the  scale  of  lengths  is  1:60  then, 

,        ,,  .  ,,  TT       28000X2087       O7n  ^  nnita 

for    deflections    25    times    actual,    this  pole    becomes   #=744x60x25  =270-° 

When  /  is  variable  the  pole  is  made  variable  as  was  done  in  Fig.  39A,  or  the  w  loads  may 

be  divided  by  /  as  is  done  in  Art.  47. 

The  force  and  equilibrium  polygons  are  now  drawn,  using  any  convement_scale  for 
the  w  forces  and  the  same  scale  for  H .  This  gives  the  deflection  polygon  A'C',  Fig. 
46Ab  with  ordinates  25  times  actual.  Thus  the  actual  end  ordinate  ^  =  1.96/25  inches. 
This 'same  polygon  is  also  the  Xa  influence  line  with  a  factor  p  =  l/i)B  irrespective  of 
the  scale  of  ordinates,  since  all  ordinates  rjm  and  TJB  are  measured  with  the  same  scale. 
To  obtain  the  Xa  influence  line  with  a  factory  1,  divide  all  the  ordinates  of  Fig.  46Ab 
by  the  end  ordinate  ,Bf  whence  the  curve  A'C',  in  Fig.  46AC,  is  obtained  with  the  end 
ordinate  WC'  =  l,  drawn  to  any  convenient  scale  of  ordinates.  The  redundant  react 
X  can  thus  be  found  for  any  case  of  concentrated  loads  P  and  is  Xa  =  ^Prl/t)B. 

To  obtain  the   Mn  influence  area   for  moments  about  the  point  n,  the    Xa  influence 
line  is  now  combined  with  the  M0/Ma  influence  line,  with  a  factor  p  =  Ma  according  t 

q'  In  Fio-  b  the  line  ^C7  represents  the  W0  influence  line  provided  the  end  ordinate 
WC'  is  made  equal  to  z,  but  this  end  ordinate  is  actually  ,fl>  hence  the  linem  question 
must  have  a  factor  p=z/yB.  But  since  Ma=z,  therefore,  the  same  line  n  C  becomes 
the  W0/Ma  influence  line  with  a  factor  p-lfa  which  is  the  same  as  the  factor  for 
the  Xa  line.  Hence  the  area  IVC7  represents  the  Mn  influence  area  with  a  fad 

/'  =  ^irnifarly/?n5'Fig.  c,  the  area  A^V  represents  the  Mn  influence  area  with  the  factor 
a=Ma=z  because  ,'fl  was  here  made  equal  to  unity.  This  was  done  to  show  the  two 
solutions  and  to  illustrate  the  fact  that  the  second  step  of  drawing  the  Z  influence  line 
with  a  =  1  is  really  unnecessary  and  adds  nothing  except  to  simplify  the  final  factor. 

The  algebraic  sign  of  the   Mn  area  is  derived  from_Eq.  (46B)  since  the  larger  _ X 
area  (which  is  positive)  is  subtracted  from  the  positive  M0/Ma  area,  leaving  a  negat 
area  as  the  remainder. 


142 


KINETIC  THEORY   OF  ENGINEERING  STRUCTURES 


Oi- 


o/Q,  LINE. 


C' 


FIG.  46A. 


ART.  47  SPECIAL  APPLICATIONS  OF  INFLUENCE  LINES  143 

The  moment  area  Mm,  for  moments  about  a  point  m,  is  indicated  to  show  a  case 
where  a  portion  of  the  area  below  the  Xa  line  is  positive,  creating  a  load  divide  at  the 
point  i. 

The  moment  area  MA,  for  moments  about  the  abutment  A,  is  also  shown.  It  is 
the  area  between  the  line  A'C'  and  the  Xa  line,  which  is  entirely  a  positive  area. 

In  all  of  these  moment  influence  areas,  the  moments  obtained  will  be  expressed  in 
the  same  units  as  the  applied  loads  P  and  the  ordinate  2.  Thus  for  P  in  Ibs.  and  z  in 
feet  the  moments  will  be  ft.  Ibs.  The  scale  of  the  ordinates  y  is  immaterial  and  is 
eliminated  in  the  factor  p,  Thus  if  Fig.  b  is  used,  the  moment  about  n  for  any  train 
of  loads  would  be 

ii;--—  S 

and  using  Fig.  c  it  would  be 


and  the  units  would  depend  solely  on  those  chosen  for  z  and  the  loads  P. 

To  obtain  the  Qn  influence  area  for  shear  at  the  point  n,  the  Xa  influence  line  is 
now  combined  with  the  Q0/Qa  line,  with  a  factor  /*=Qa  =  l,  according  to  Eq.  (46e), 
see  Fig.  46Ad. 

For  a  load  unity  to  the  right  of  n,  the  shear  in  the  principal  system  would  be 
Q)  =  l  and  for  Xa  =  l  the  shear  Qa  =  l,  hence  Q0/Qa  =  ^  so  that  the  broken  line  A'n'DC' 
represents  the  Q0/Qa  influence  line  with  a  factor  unity,  and  the  Qn  influence  area  is  the 
shaded  area  with  a  factor  /*=Qa  =  l,  with  the  portion  below  the  Xa  line  positive,  as 
before.  Were  Fig.  b  used,  the  factor  would  be  /w=Qa/'?B==l/'?s- 

The  point  n  is  always  a  load  divide  for  QH. 

The  end  reaction  A  for  any  system  of  loads  is  simply  HP  —  Xa,  from  Eq.  (46B), 
because  A0  =  DP  and  Aa  =  l. 

A  uniform  live  load  covering  the  whole  span  would  give  Xa  =3pl/8,  as  may  be  found 
from  Eq.  (15j).  The  graphic  diagram,  Fig.  46Ab,  gave  Xa=7Al2p=0.372pl 

ART.  47.     PLATE    GIRDER   ON   THREE   SUPPORTS 

The  example  chosen  is  a  general  case  of  unequal  spans  and  variable  moment  of 
inertia  as  illustrated  in  Figs.  4?A.  All  dimensions  in  inches,  and  the  loads  are  directly 
applied  to  the  girder. 

According  to  the  criterion  of  Eq.  (3c)  this  structure  involves  one  external  redundant 
condition.  Any  one  of  the  three  supports  may  be  taken  as  the  redundant  condition, 
but  it  is  generally  most  convenient  to  assume  the  middle  support  at  C  and  thus  obtain 
the  principal  system  as  a  simple  beam  on  two  supports  A  and  B. 

The  Xc  influence  line  for  the  redundant  support  is  again  given  by  Eq.  (42n)  as 


(47A) 


where  Smc  is  the  variable  influence  line  ordinate  at  any  point  m  with  the  constant  factor 
fi  =  \/dect  and  dmc  and  dcc  must  be  measured  to  the  same  scale.     The  ordinates  dmc  are 


144  KINETIC  THEORY  OF   ENGINEERING  STRUCTURES          CHAP,  x 

the  ordinates  of  a  deflection  polygon  drawn  for  the  conventional  loading  Xc  =  \  kip, 
acting  downward  at  C  on  the  principal  system  or  simple  beam  AB. 

This  deflection  polygon  is  the  equilibrium  polygon  drawn  for  a  series  of  elastic  loads* 
w  =  MJx/I  when  the  pole  H=E.  The  variable  moment  of  inertia  is  not  taken  care  of] 
in  the  manner  previously  given,  by  varying  the  pole  distance,  but  the  alternative 
method  of  involving  /  in  the  w  loads  is  here  employed.  The  moments  of  inertia  are 
given  for  the  various  sections  of  the  beam  in  Fig.  4?Aa.  The  M0  diagram  for  Xc  = 
1000  Ibs.  is  drawn  in  Fig.  47Ab,  with  an  ordinate  under  the  point  C  equal  to  1000  l^lo/l  = 
175,000  in.  Ibs. 

The  several  M  ordinates  are  scaled  from  the  diagram  or  computed,  and  these 
values  divided  by  the  corresponding  7's  furnish  the  figures  for  plotting  the  ~kZ/7  diagram^ 
Thus  the  ordinate  at  C  is  1  75000  H-  34982  =5.  00.  The  w  loads  are  then  computed  aa 
the  areas  of  the  M/l  diagram  using  the  horizontal  distances  between  the  ordinates  in 
inches.  These  loads  are  applied  at  the  centers  of  gravity  of  the  respective  areas. 

The  w  loads  are  now  combined  into  a  force  polygon,  Fig.  47AC,  with  pola 
H  =El  100  XI  20  -=2333  w  units,  all  drawn  to  the  same  scale,  but  this  scale  may  be  taken 
as  any  convenient  one  without  regard  to  the  deflections.  The  particular  pole  will  then] 
give  deflections  one  hundred  times  actual,  because  the  scale  of  lengths  for  the  girder 
span  was  chosen  1:120  and  the  pole  was  made  100X120  =  12000  times  too  small. 
I/  =28,000,000  Ibs.  per  sq.  inch.  The  resulting  equilibrium  polygon  'A'C'B',  Fig.  47ACJ 
is  the  deflection  polygon  for  the  load  XC  =  IQOO  Ibs.  with  ordinates  100  times  actual 
and  it  is  also  the  Xc  influence  line  with  a  factor  1/c  where  c  is  the  ordinate  under  C. 
measured  to  the  same  scale  as  the  other  influence  ordinates. 

It  is  clear  from  Eq.  (4?A)  that  the  scale  of  the  influence  ordinates  is  immaterial  so; 
long  as  the  same  scale  is  used  for  the  ordinate  c,  but  when  actual  deflections  are  sought 
then  a  natural  scale,  as  inches,  must  be  employed. 

The  reaction  Xc  for  any  case  of  concentrated  loads  may  then  be  found  from  Fig. 
47AC,  by  multiplying  the  loads  by  the  d  ordinates  and  dividing  the  sum  of  these  products 
by  c,  thus  Xc  =  ^Pd/c,  where  c  and  the  d's  are  measured  to  the  same  convenient  scale. 

The  A  and  B  influence  areas  are  easily  found  when  the  Xc  line  is  given.  Thus  in 
Fig.  47  AC,  the  line  B'C'A"  represents  the  A0/Aa  influence  line  with  some  factor,  and 
the  shaded  area  is  the  A  influence  area  with  a  factor  I/TJA  to  be  proven. 

For  this  proof  the  end  ordinate  A'  A"  is  evaluated  according  to  Eq.  (46s),  and  the 
factor  required  for  this  ordinate  will  be  the  factor  for  all  ordinates  TJ  of  the  A  influence 
area. 

For  the  point  A,  A0  =  l  and  Ac  =  l-l2/l  and  Xc  has  the  factor  1/c.  Hence  ~A0f  Ac  = 
1/12  and  the  end  ordinate  should  be 

i2\i    xc-\    i2\d      I 
=~=  ''   ......    (7B) 


where  the  Tactor  is  l2/cl. 

But  —=—  -  =  —  ,  hence  the  factor  is  simplv  \fr\A. 

d         C         T)A' 


ART.  47 


SPECIAL  APPLICATIONS  OF  INFLUENCE  LINES 


145 


k               9i 

rlW 

P 
•n       t                    5 

1=  fZ.U 
6 

7 

a       3 

10 

u 

i 
i 

0                 1 

1 

3 

S.94 


*u= 


FIG.  47x. 


146  KINETIC  THEORY  OF    ENGINEERING  STRUCTURES  CHAP.X 

. 
Similarly  \/rIB  is  the  factor  for  the  B  influence  area  indicated  in  Fig.  47 AC  as  the 

area  included  between  the  Xc  line  and  the  line   A'C'B". 

Hence  the  reactions  A  and  B,  for  any  train  of  concentrated  loads,  become 

A=  —  2Pry       and       B=  —  2Pri'.  (47  c) 

f)A  "TiB  .-. 

The  algebraic  signs  of  the  areas  are  determined  as  before.  All  areas  below  the  Xc 
line  are  positive  and  those  above  this  line  are  negative.  The  ordinates  may  be  measured 
with  any  scale  so  long  as  the  same  scale  is  used  for  all. 

The  M  influence  area  is  a  portion  of  the  A  area  for  all  points  between  A  and  (7, 
while  for  points  between  C  and  B  the  M  area  is  a  portion  of  the  B  area.  See  Fig. 
4?Ad. 

Thus  for  the  point  4  the  M4  influence  area  is  shown  as  the  shaded  area  between 
the  Xc  line  and  the  broken  line  A'-i'B'.  The  factor  is  MC/C=X/TJA,  where  x  is  the 
ordinate  of  the  point  4.  When  x  is  expressed  in  feet  the  resulting  moment  is 
in  ft.  Ibs. 

Similarly  the  M-j  influence  area  is  indicated  by  dotted  lines  and  the  factor  is 
Mc/c=x/r)B. 

The  usual  directions  as  to  algebraic  signs  and  scales  apply  as  before. 

The  Q  influence  area  for  shear  is  shown,  in  Fig.  4?Ae,  for  the  point  4  as  the  shaded 
area  with  a  factor  QC/C  =  \/T)A  the  same  as  for  the  A  influence  are.a.  The  line  A'47  is 
parallel  to  B'C'4". 

Similarly  the  Q7  influence  area  is  indicated  by  dotted  lines  and  has  a  factor  Qc/c  = 
I/T)B'  B'7'  is  parallel  to  A'C'7".  The  signs  are  again  determined  with  reference  to 
the  areas  above  and  below  the  Xc  line. 

All  of  these  influence  areas  have  a  load  divide  at  the  point  C'  and  the  Q  areas  have 
two  load  divides  each.  This  determines  the  positions  of  train  loads  for  positive  and 
negative  effects,  and  to  obtain  maximum  effects,  the  heaviest  loads  should  be  placed 
over  the  maximum  ordinates. 

Temperature  effects.  A  uniform  change  in  temperature  will  expand  the  girder 
equally  in  all  directions  and  will  produce  a  slight  lifting  of  the  ends  A  and  B  equal  to 
d(72- 12)  =0.0105  inch,  for  £=25°  F.  and  £=0.000007.  When  the  three  supports  are 
on  the  same  level  then  no  stress  will  be  produced  by  uniform  temperature  changes. 

However,  when  the  sun  shines  down  on  such  a  structure  experience  teaches  that 
the  top  flange  is  heated  much  more  than  the  bottom  flange  and  this  difference  in  tem- 
perature may  become  quite  considerable  and  will  cause  the  girder  to  assume  a  curved 
position,  convex  upward,  when  the  top  flange  is  warmer  than  the  bottom. 

The  maximum  difference  in  temperature  between  the  two  flanges  is  bound  to  remain 
somewhat  problematic,  but  observations  indicate  that  differences  of  30°  F.  are  quite 
common.  In  the  present  example  Jf=+20°  F.  will  be  assumed  and  on  this  basis  and 
that  the  temperature  varies  uniformly  between  the  two  flanges,  a  set  of  wt  elastic  loads 
is  now  computed  from  the  formula  wt=  -sJUx/h  =  -0.00014 dx/h,  where  Ax  is  the 
horizontal  distance  between  sections  and  h  is  the  depth  of  girder,  both  in  inches. 


AKT.  47  SPECIAL  APPLICATIONS  OF  INFLUENCE  LINES 

The  following  table  gives  the  wt  loads: 


147 


Point. 

dx 
Inches. 

h 

Inches. 

»l 

10,000u>( 

Point. 

dx 
Inches. 

h 
Inches. 

wt 

10,000u>, 

1 

60 

19.5 

0.00043 

4.3 

7 

60 

72 

0.000117 

1.17 

') 

60 

34.5 

0.00024 

2.4 

8 

60 

72 

0.000117 

1.17 

3 

60 

49.5 

0.00017 

1.7 

9 

45 

64.5 

0.000098 

0.98 

4 

60 

64.5 

0.000115 

1.15 

10 

45 

49.5 

0.000127 

1.27 

5 

90 

72 

0.000175 

1.75 

11 

45 

34.5 

0.000183 

1.83 

6 

90 

72 

0.000175 

1.75 

12 

45 

19.5 

0.000323 

3.23 

An  equilibrium  polygon  drawn  for  these  wt  loads  with  a  pole  distance  unity,  would 
represent  the  deflection  curve  of  the  beam  due  to  M.  However,  for  convenience  it  is 
better  to  multiply  these  small  loads  wt  by  some  factor  as  10,000  and  then  make  the 
pole  distance  10,000  instead  of  unity. 

The  scale  of  lengths  for  the  girder  was  made  1 : 120  and  if  the  ordinates  are  to  appear 
say  5  times  actual,  the  pole  distance  should  be  made  equal  to  H  =  10,000/5X120  = 
16.67,  using  the  scale  chosen  for  the  wt  loads. 

Fig.  47Af  shows  the  force  and  equilibrium  polygons  for  the  wt  loads,  with 
ordinates  five  times  actual.  Since  these  wt  loads  are  all  negative  for  +At,  the 
3ct  deflection  curve  was  drawn  above  the  closing  line  A  'B'.  The  ordinate 
under  the  point  C  represents  5dct  and  this  must  be  increased  by  the  small 
ordinate  5x0.0105,  previously  found  for  the  uniform  rise  in  temperature  to  obtain 
total  effect. 

According  to  Eq.  (44c),  the  redundant  reaction  produced  by  this  temperature 
effect  is  Xct=dct/dee.  The  deflection  polygon  for  Xe  =  l  kip,  gives  100£cc=0".93, 
whence  +  £cc=0".0093,  and  the  dct  deflection  polygon  gives  -5dc<=0".74+0.05=0".79, 
making  3ct  =  — 0'M6. 

Hence  Xct  =det/dce  =  -0.16/0.0093  =  -17.2  kips,  which  indicates  a  downward  reaction 
at  C  to  maintain  the  girder  on  the  middle  support.  As  the  entire  girder  has  an  approxi- 
mate weight  of  7440  Ibs.  the  above  temperature  effect  would  lift  the  girder  off  the  center 
support  and  cause  the  beam  to  carry  itself  on  two  supports,  producing  a  compression 
of  about  3100  Ibs.  per  sq.in.  on  the  extreme  fiber.  This  temperature  stress  would  not, 
however,  be  fully  developed  unless  the  span  is  loaded  down  in  contact  with  the  center 
support. 

The  dead  and  live  load  stresses  must  finally  be  combined  with  the  temperature 
effect  to  obtain  the  real  stress  at  any  point. 

The  application  of  these  influence  areas  is  illustrated  by  placing  a  single  load  Pn  at 
any  point  n  of  the  span  and  showing  the  values  of  the  several  functions.  The  same 
process  is  followed  for  each  load  of  a  train  of  loads  and  the  total  effect  of  the  train  is 
the  sum  of  the  individual  load  effects. 

The  ordinates  r?  under  the  point  n,  in  each  of  the  influence  areas,  are  all  scaled  to 
the  same  scale  to  which  7j_4  and  rjB  are  measured.  This  may  be  any  convenient  scale, 
which  is  universally  used  for  all  ordinates  of  the  drawing. 


148  KINETIC  THEORY   OF    ENGINEERING  STRUCTURES  CHAP.  X 

The  values  of  the  various  functions  for  the  load  Pn  are 

Y          fnOn  _  '  -°p     _ 

Ac=  --  —  /"„  — 

C  57.  0 


B  =  _M  =  _-M-  PB  =  -0.207Pn  ; 
ijjs  15.94 

+-B=P,  which  checks  to  1  percent. 

n  =49.4Pn  in.lbs.  for  Pn  Ibs. 


Xctl2a     17200X300X194  '. 

M4t  —  J2-  =  -          72Q  =1,390,330  in.lbs. 


ART.  48.     TRUSS   ON  THREE   SUPPORTS 

Figs.  48A  illustrate  the  analysis  by  influence  lines  of  a  truss  on  three  supports. 
The  bottom  chord  is  the  loaded  chord  and  the  lengths  and  cross-sections  of  the  members 
are  written  over  the  members  in  Fig.  a,  while  the  stress  SA,  produced  by  a  unit  reaction 
at  A  on  the  simple  span  AB,  are  written  below  the  members.  The  stress  diagram, 
from  which  the  SA  stresses  are  found,  is  shown  in  Fig.  b.  The  solution  is  similar  to 
that  given  in  Art.  47. 

As  in  the  previous  problem,  Art.  47,  the  present  structure  involves  one  external 
redundant  condition,  which  is  again  taken  as  the  middle  support  C,  reducing  the  truss 
to  a  principal  system  on  two  supports,  A  and  B. 

The  Xc  influence  line  is  derived  from  Eq.  (42o)  as  in  the  previous  problem  and 
becomes 


(48A) 


where  the  variable  influence  line  ordinate  dmc  is  the  ordinate  for  a  deflection  polygon, 
drawn  for  the  loaded  chord  and  for  the  conventional  loading  Xc  =  1  kip  acting  downward 
at  C  on  the  prinicpal  system  or  simple  truss  AB.     The  influence  line  factor  ju  =  l/£cc. 
The  elastic  loads  for  this  deflection  poylgon  are  found  from  Eqs.  (36s)  as  : 

Mc  I 

•  •  -  •  (48B) 


for  each  pin  point  of  the  top  and  bottom  chords.  The  effect  of  the  web  members  is 
usually  neglected  as  being  insignificant,  but  when  it  is  desired  to  include  their  effect, 
Eqs.  (36c)  will  give  the  w  loads  due  to  the  web  members  and  these  are  added  to  the 


ART.  48 


SPECIAL  APPLICATIONS  OF  INFLUENCE  LINES 


149 


chord  loads  found  from  Eq.  (48R).  See  Art.  36,  for  a  complete  discussion  of  this  method, 
and  Art.  50  for  a  complete  problem. 

A  Williot-Mohr  diagram  might  also  be  employed  to  obtain  this  deflection  polygon. 
See  also  the  example  in  Art.  50,  Fig.  50B. 

In  the  present  example  the  web  system  will  be_  neglected  and  the  w  loads  arc  found 
for  the  chords,  using  Eq.  (48 B).  The  moments  M  are  those  produced  by  the  conven- 
tional loading  Xc  =  1000  Ibs.  and  are  represented  in  the  moment  diagram,  Fig.  b,  expressed 
in  inch-lbs.  The  ordinate  at  the  point  C  will  be  P//4  =528,000  in.  Ibs. 

Using  these  moments  and  the  values  for  /,  F  and  r  given,  in  Fig.  a,  for  each  member, 
the  following  computation  of  the  w  forces  is  made: 


w 

Member. 

A/c 
Inch-lbs. 

I 
Inch. 

F 
Square  Inches. 

r 
Inch. 

Md 
Fr* 

1 

L0L2 

127,125 

508.5 

22.7 

384 

14.8 

2 

t/!^ 

254,250 

509.0 

42.0 

384 

20.8 

3 

L2L4 

381,375 

508.5 

22.7 

384 

57.7 

4 

C73f75 

528,000 

299.3 

44.7 

434.9 

2X18.7 

5 

L4L6 

381,375 

508.5 

22.7 

384 

57.7 

6 

£75t/7 

254,250 

509.0 

42.0 

384 

20.8 

7 

L6L8 

127,125 

508.5 

22.7 

384 

14.8 

* 

z». 

224.0 

The  modulus  #=29,000,000  /6s.  per  xq.  inch,  was  not  embodied  in  the  tabulated 
values,  hence  the  w  loads  are  E  times  too  large  and  the  deflections  would  be  natural  size 
for  a  pole  H=E.  But  the  scale  of  lengths  of  the  drawing  was  1:300,  and  wishing  to 
make  the  deflections  400  times  actual  the  pole  must  be  made  equal  to  E/3QQ  X400  =241. 1 'w 
units. 

The  force  and  equilibrium  polygons,  Fig.  c,  are  then  drawn,  using  any  convenient 
scale  for  w  forces.  The  Xc  influence  line  is  thus  found  as  the  polygon  A CB,  with  a 
factor  ft  =  l/c.'  The  actual  deflection  of  the  point  C  is  dee  =  l". 07/400=0". 0027, 
obtained  by  measuring  the  ordinate  c  in  inches.  All  influence  ordinates  are  measured 
with  the  same  scale,  which  may,  however,  be  any  convenient  scale,  because  the  factor 
JJL  makes  the  quotients  dmc/dcc  independent  of  an  absolute  scale.  For  this  same  reason 
the  M  diagram  might  have  been  drawn  with  any  middle  ordinate  as  unity,  though 
when  the  actual  deflection  dcc  is  wanted  for  temperature  investigations,  the  method 
here  given  is  less  confusing. 

The  A  influence  area  is  now  found  by  combining  the  Xc  influence  line  with  the 
ordinary  A  line  for  the  span  AB,  in  such  manner  as  to  make  the  ordinate  at  C  equal 
to  zero  and  then  finding  the  factor  u  which  reduces  the  ordinate  at  A  to  unity. 

The  line  A'CB,  Fig.  c,  is  the  ordinary  A  influence  line  with  a  factor  I/IJA,  and 
when  this  is  combined  with  the  Xc  influence  line,  the  areas  included  between  the  two  lines 
will  represent  the  A  influence  area  with  a  factor  ^  =  !/TJA  =0.0187. 

Since  it  is  more  difficult  to  redraw  the  Xc  line  so  as  to  make  the  ordinate  c  =  l, 
than  it  is  to  apply  the  factor  p,  the  latter  method  is  decidedly  preferable  to  the  one 
frequently  given. 


150 


KINETIC  THEORY  OF  ENGINEERING  STRUCTURES 


CHAP.  X 


t  i         ?  1 

I  A    INFLUENCE    AREA  ,  |fX»/rk=»+0.|Ql87 


INFLUECE    AREA  , 


Upper  chord 
I  Lower  chord 


Tension  is  •+•  ,  compreision  is  — 


ART.  48  SPECIAL  APPLICATIONS  OF  INFLUENCE  LINES  151 

The  point  C  is  a  load  divide  for  both  reactions  A  and  B,  and  in  the  present  case 
the  same  influence  area  will  serve  for  both  these  reactions,  because  l\  =12.  When  the 
spans  are  unequal,  then  the  B  influence  area  is  included  between  the  Xc  line  and  the 
line  AC  prolonged  to  the  vertical  through  B  and  the  factor  then  becomes  ,«B  =  I/>?,B. 

The  influence  area  for  a  chord  member  L2L4  is  obtained  by  combining  the  ordinary 
S0  stress  influence  line  of  the  member  for  the  principal  system  with  the  A"c  influence 
line,  Fig.  d. 

A  load  at  C  can  produce  no  stress  in  any  member  of  the  indeterminate  truss,  hence 
the  ordinate  under  C,  for  every  stress  influence  line,  must  be  zero.  Therefore,  the  line 
BCA'  must  be  common  to  all  stress  influence  lines  of  the  members  between  A  and  C, 
web  members  included.  The  point  C  must  be  a  load  divide  for  all  members. 

The  center  of  moments  for  the  chord  L2L4  is  at  C73,  hence  the  line  A3,  Fig.  d, 
completes  the  S0  stress  influence  line  with  a  factor  SA/T)A,  where  SA  is  the  stress  in  the 
chord  L2L4  for  a  unit  reaction  at  A. 

The  area  included  between  the  Xc  polygon  and  the  S0  influence  line  is  thus  the 
required  influence  area  for  the  chord  L2L4  with  the  factor  JJ.=SA/^A-  The  area  below 
the  Xc  polygon  is  positive  as  usual.  The  stress  *SU  =  +1.99,  hence  the  factor 
/i  =+1.99/53.6  =+0.0371. 

If  the  vertical  L3t/3  were  absent,  then  the  S0  line  would  have  to  be  straight  over 
the  panel  L2Z/4  and  a  line  2C  would  be  necessary  to  complete  the  S0  influence  line. 
With  the  member  L3t/3  acting,  the  chord  stress  L2L4  is,  therefore,  greater  than  when 
the  vertical  is  omitted. 

The  S0  influence  lines  and  their  factors  for  all  chord  members  are  shown  by  dotted 
lines  in  Fig.  d,  and  the  final  influence  areas  are  always  positive  below  the  Xc  line.  The 
factors  ft  are  negative  for  the  top  chords  because  SA  is  then  negative. 

The  stress  in  the  member  L2L4,  for  any  train  of  moving  loads,  is  expressed  by 


(48c) 


where  each  ij  is  the  ordinate  of  the  shaded  influence  area  vertically  under  its  lespective 
load  P  and  SA  is  given  in  the  same  units  as  the  loads. 

This  influence  area  is  also  the  M3  influence  area  with  a  factor  fi=x3/i)A  and  gives 
moments  in  foot  units  when  x3  is  measured  in  feet  as  indicated. 

The  scale  of  the  influence  ordinates  is  immaterial  so  long  as  all  ordinates  are 
measured  with  the  same  scale.  This  is  apparent  from  Eq.  (48A). 

The  influence  area  for  a  web  member  L2U3  is  found  from  exactly  similar  considerations 
as  those  shown  to  exist  for  the  chords.  The  shaded  area  in  Fig.  e  is  the  influence  area 
for  the  web  member  L2U3.  The  S0  lines  for  the  other  web  members  are  indicated  by 
dotted  'lines. 

The  \me-BCA',  Fig.  e,  is  again  one  of  the  limiting  lines  of  the  S0  influence  line  for 
each  web  member,  and  when  the  chords  are  parallel  in  the  panel  containing  the  particular 
web  member,  then  the  other  -limiting  line  A2'  will  be  parallel  to  BC  and  the  line  2'3 
completes  the  S0  line  for  the  member  L2US. 

The  load  divide  i  for  this  member,  considering  the    principal  system,  must  fall 


152  KINETIC   THEORY  OF  ENGINEERING   STRUCTURES          CHAP,  x 

vertically  over  the  point  i'  where  the  line  2'3  intersects  the  base  line  AH.  This  serves 
as  a  check  or  when  the  chords  are  not  parallel,  as  in  the  panel  L3L4,  then  it  may  be 
used  to  complete  the  S0  line  as  indicated  for  the  member  t/3Z/4  with  the  load  divide  t. 

In  case  the  chords  are  not  parallel  and  the  center  of  moments  for  the  diagonal 
falls  outside  the  drawing,  then  the  ordinate  753  as  for  the  member  USL4,  may  be  com- 
puted from  the  formula 

r/3      x3+a3  Jx3+a3\  /63.6+91\.Q 

u-ir   or  ^3=t^>4  =  r^r-r3-6<  •  •  •  ^ 

where  x3  is  the  distance  L0L3  from  A  to  the  panel  containing  the  web  member,  and  as 
is  the  distance  from  A  to  the  center  of  moments  of  the  member,  both  in  the  same  units, 
as  feet. 

The  factor  {J.=SA/T)A  again  applies  and  the  sign  of  SA  determines  the  sign  for  //. 

Stresses  due  to  temperature.  When  the  three  supports  A,  B  and  C  are  on  the  same 
level  then  a  uniform  change  in  temperature  produces  no  stresses  in  the  structure. 
However,  when  the  sun  illuminates  the  bridge  from  above,  there  is  usually  a  difference 
in  the  temperature  of  the  two  chords.  Assuming  this  difference  as  Jf=20°  F.  as  in 
Art.  47,  then  the  bottom  chord  would  be  cooler  than  the  top  chord  by  this  amount  and 
the  changes  in  the  lengths  of  the  bottom  chord  members  would  be  —  eMl  and  the  wt 
elastic  loads  would  be 


(48E) 


These  loads  being  extremely  small,  it  is  well  to  multiply  them  by  say  10,000  and 
then  using  a  pole  distance  of  10,000,  the  dcf  deflections  would  be  natural  size.  However. 
the  scale  of  lengths  on  the  drawing  was  1  :  300  and  for  deflections  five  times  natural  the 
pole  becomes 

10000      rr_ 
=5X300  *  units. 

The  loads  wt  are  figured  for  the  panel  points  1,  3,  5  and  7,  taking  £=0.000007, 
J*=20°  F.  and  r=384  inches.  The  dct  deflection  polygon  is  shown  in  Fig.  d,  and  has 
negative  deflections,  five  times  actual.  Hence  oct  =0".95/5  =0".19  and  dcc  was  previously 
found  from  Fig.  e,  as  0".0027,  hence  from  Eq.  (44c) 


This  value  is  in  kips  because  dcc  is  the  actual  deflection  for  one  kip,  and 
Xct  being  negative  produces  the  same  effect  on  the  principal  system  as  a  load  Xct  hung 
at  the  point  C.  The  reaction  at  A,  due  to  a  load  of  70.4  kips  at  C,  would  then 
be  At  =70.4Z2/(Zi  +Z2)  =  +35.2  kips. 

The  stresses  in  the  members  are  then  St=SAAt=35.2  SA  kips.  The  stresses 
SA,  due  to  a  unit  reaction  at  A,  are  already  found  in  Fig.  b,  hence  the  tem- 
perature stresses  are  readily  determined.  Thus  for  the  member  U3U4,  SA  =  -2A2, 
hence  St  =  -2.42  X35.2  =  -85.2  kips.  The  negative  sign  indicates  compression. 


ART.  49 


SPECIAL  APPLICATIONS  OF  INFLUENCE  LINES 


153 


ART.  49.     TWO-HINGED   SOLID   WEB   ARCH   OR   ARCHED   RIB 

This  and  the  two-hinged  framed  arch  are  perhaps  the  most  common  structures 
involving  external  redundancy  which  are  met  with  in  practice.  They  will  receive 
special  attention  here  as  deserving  a  prominent  place  among  commendable  structures. 
The  external  redundancy  may  be  said  to  offer  less  objection  here  than  in  any  other 
class  of  structures. 

The  present  theory  will  be  developed  in  its  most  general  application  to  unsym- 
metric  arches  and  will  be  equally  applicable  to  masonry,  concrete  or  steel  arched  ribs. 

Fig.  49A  represents  an  unsymmetric  two-hinged  arched  rib  of  any  cross-section  and 
the  lettered  dimensions  are  in  general  the  same  as  those  previously  employed  for  three- 
hinged  arches  in  Art.  28. 

The  arch  thrust  along  AB  is  treated  as  the  external  redundant  condition.  When 
it  is  removed,  by  replacing  the  hinged  support  at  A  by  a  roller  bearing,  the  simple  beam 
AB  (though  curved)  on  two  supports,  then  becomes  the  principal  system. 


FIG.  49A. 


According  to  Eqs.  (7  A),  the  following  values  for  the  reactions,  thrusts  and  moment 
for  any  point  ra  may  be  written: 


A  =A0  —AaXa,       where 
B=B0-BaXa,       where 


and       Aa=-l-sina 


and 


=  l-  since 


Mm=M0-MaXa,     where       M0=A0x-P(a'  -x')     and     Ma  =  l-ycosa 

Nm  =  N0  -  NaXa,      where       N0  =  Q0  sin  <f>  ;       Na  =  -  1  •  cos  (<f>  -a) 

Tm  =  T0-TaXa,    where   T0=Q0cos<f>  =  (A0-Xp)  cos<£;     77«  =  l-sin(0-a) 


(49A) 


154  KINETIC  THEORY  OF   ENGINEERING  STRUCTURES  CHAP.  X 

in  which  A  and  B  are  the  vertical  end  reactions;  Mm  is  the  moment  of  the  external 
forces  about  any  axial  point  m;  Rm  is  the  resultant  thrust,  on  the  normal  section  mm', 
produced  by  all  the  external  forces  acting  on  one  side  of  this  section;  Nm  is  the  com- 
ponent of  Rm,  normal  to  the  section  mm';  and  Tm  is  the  tangential  component  of  Rm 
in  the  section  mm'.  Q0  is  the  vertical  shear  at  any  point  m  produced  by  all  the  loads 
P  acting  on  a  simple  beam  AB.  The  conventional  loading  Xa  =  \  is  applied  as  indicated 
in  the  opposite  direction  of  Xa. 

Eqs.  (49A),  with  the  special  values  inserted,  thus  become 

SPa'  . 

A  =  —  ,  --  1-  Xa  sm  a 


v     . 
Xa  sm  a 


I 

Mm  =M0  —Xay  cos  a 
Nm  =Q0  sin  </>  +Xa  cos  (<£  -a) 
Tm  =Q0  cos  <£  —Xa  sin  (<f>  -a)  J 


/ 

SPa 

— A,,  sm  a 

(49s) 


where  M0  and  Q0  have  the  values  in  Eqs.  (49A). 
From  Fig.  49A, 


Mm=Nmv;      Rm=VNm2  +  Tn?;       H=Xacosa.     ^     .     .     . 

In  the  above  equations,  all  terms  except  Xa  are  derived  from  the  static  conditions 
and  the  solution  becomes  possible  when  Xa  is  determined. 

The  Xa  influence  line  for  the  redundant  haunch  thrust.  When  the  haunches  are  on 
the  same  level  then  this  thrust  Xa  becomes  the  horizontal  thrust  H. 

The  equation  for  Xa  is  given  by  Eq.  (42o)  as 


where  dma  is  the  vertical  deflection  ordinate  of  any  point  m  of  a  deflection  polygon, 
drawn  for  the  loaded  chord  and  for  the  conventional  loading  Xa  =  l  applied  to  the 
principal  system,  and  o^  is  the  change  in  the  length  AB,  due  to  Xa  =  \  acting  on 
the  principal  system. 

Since  dma  and  d^  are  not  parallel  displacements,  they  cannot  be  obtained  from 
the  same  deflection  polygon  as  in  the  previous  examples,  Arts.  46^8.  This  circum- 
stance necessitates  the  construction  of  an  extra  displacement  polygon  for  daa,  if  the 
graphic  solution  is  strictly  followed,  or  of  finding  d^a  by  computation,  which  is  usually 
advisable. 

The  Xa  influence  line  thus  becomes  the  dTOa  deflection  polygon  with  a  factor  p.  =  l/§aa- 
Should  this  deflection  polygon  be  constructed  for  a  pole  distance  H  =daa  then  the  factor 

p-1. 

According  to  Art.  40,  the  dma  deflection  polygon  is  the  equilibrium  polygon  drawn 
for  a  set  of  elastic  loads  w  with  a  pole  distance  H  =  1.  These  elastic  loads  represent 
partial  areas  of  a  moment  diagram  drawn  for  the  conventional  loading  Xa  =  l. 


ART.  49  SPECIAL  APPLICATIONS  OF  INFLUENCE  LINES  155 

The  moment  produced  by  Xa  =  1  about  any  axial  point  m  is 

M,n  =  1  •  y  cos  a       ..........     (49E) 

and  the  elastic  loads,  by  Eq.  (39i)  become 

M  Au     Axy  cos  a 


^-JT--J—  --  -r, 

El       El  cos  (j>  ' 


(49F) 


where  Au  is  the  width  of  a  partial  moment  area  measured  along  the  axial  line,  and  Ax 
is  the  horizontal  distance  between  the  vertical  moment  ordinates,  making  Au=4x/cos<j>. 

The  deflection  daa  may  be  obtained  from  a  deflection  polygon  drawn  for  the  same 
set  of  w  loads  by  allowing  these  loads  to  act  in  a  direction  parallel  to  Xa,  as  per  Art.  38. 

According  to  Eq.  (38B)  the  deflection  of  any  point  in  any  direction,  may  be  expressed 
as  the  sum  of  the  moments  of  all  w  loads  on  one  side  of  the  point,  when  the  direction  of 
the  w  loads  is  taken  parallel  to  the  required  direction  of  the  deflection. 

Hence 


a  daa    =  ^T    ^(J/  COS  «) 


(49o) 


which  affords  a  purely  analytic  solution  for  computing  the  Xa  influence  line  ordinates 
and  also  the  dm  displacement. 

Since  the  w  loads  are  best  found  by  computation,  it  is  a  comparatively  easy  task 
also  to  compute  the  values  wy  cos  a  and  thus  obtain  daa  =  I>wy  cos  a,  which  is  the 
required  pole  distance  for  the  dma  deflection  polygon.  Hence  according  to  Eq.  (49n) 


i 
and  the  Xa  influence  line  is  an  equilibrium  polygon  drawn  for  the  elastic  loads  w  with 

a  pole  distance  H  =  Swn/  cos  a. 

Since  E  enters  into  w,  and  hence  into  each  term  of  Eqs.  (49o),  as  a  constant  factor, 
which  cancels  in  the  numerator  and  denominator  of  Eq.  (49n),  it  would  be  proper  to 
multiply  Eq.  (49r)  by  E  and  thus  make  the  elastic  loads  equal  to  Ew  without  affecting 
the  ordinates  r?a  from  Eq.  (49n).  Should  it  be  desirable  to  obtain  values  of  dma  from 
the  Xa  influence  line,  then  the  jja  ordinates  must  be  divided  by  E  and  multiplied  by 


cos  a. 


. 

For  very  flat  arches,  the  effect  of  the  axial  thrust  should  be  considered  in  the 
determination  of  Xa,  by  allowing  for  the  quantity  dfm  produced  by  Na,  Eq.  (49A),  on 
the  displacement  daa- 

Thus  for  Na  =  -I  •  cos  (<£  -a)  and  Au  =  Jar/cos  <j>,  then  from  Eq.  (15.v), 


2  /j 

- 


156  KINETIC  THEORY  OF  ENGINEERING   STRUCTURES  CHAP.  X 

which  is  the  effect  due  to  Na  only,  and  this  added  to  the  displacement  previously  found 
for  moments  only,  gives  the  total  displacement 

2       — 

(49HH) 


Hence,  when  axial  thrust  is  to  be  considered  in  Eq.  (49n),  then  the  value  3aa  should 
be  computed  from  Eq.  (49HH)  and  used  as  the  pole  distance  H  in  drawing  the  equilibrium 
polygon  for  the  elastic  loads  w.  When  the  loads  w  are  taken  E  times  actual,  then  the 
E  in  the  second  term  of  Eq.  (49HH)  is  omitted. 

Temperature  stress.  For  any  change  in  temperature  above  or  below  the  normal, 
Xat  is  expressed  by  Eq.  (44c)  as 

Xat=m^,      .....     V     .    '.I/   ..    .     (49i) 

Oaa 

wherein  dal  is  the  change  in  the  distance  AB  due  to  any  change  in  temperature  of  ±£° 
acting  on  the  principal  system. 

When  the  structure  retains  a  uniform  temperature  then  the  change  ±t°  will  be  uniform 
in  all  directions  so  that  the  shape  of  the  structure  will  always  be  similar  to  its  normal 
geometric  shape.  Hence,  the  distance  AB  will  change  as  for  any  case  of  linear  expan- 

sion and 

dl  0.000488/ 


for  e  =0.0000065  per  1°F.,  and  Z  =  ±75°. 

When  the  change  in  temperature  is  not  uniform,  as  when  the  upper  flange  is  +Jlc 
warmer  than  the  lower  flange,  then  dat  must  be  found  from  a  deflection  polygon  drawn 
for  a  set  of  wt  elastic  loads  computed  from  the  formula  wt  =  —  edtJu/D,  where  D  is  the 
depth  of  section  and  Au  is  the  length  between  sections  measured  along  the  axis.  The 
pole  distance,  if  made  equal  to  unity,  will  give  the  actual  oat,  and  if  the  pole  be  made 
equal  to  daa  =  'Zwy  cos  a  as  for  Xa,  then  the  displacement  found  will  be  Xnt  measured 
parallel  to  AB.  The  wt  loads  must  likewise  be  taken  parallel  to  AB  in  drawing  the  force 
and  equilibrium  polygons  since  dat  is  a  displacement  along  AB. 

Xat  will  be  positive  for  +t,  while  for  +  At  the  wt  loads  are  negative  and  Xat  would 
also  be  negative. 

In  any  case,  when  Xat  is  determined,  then  the  quantities  Mmt,  Nmt,  and  Tmt  are 
found  from  Eqs.  (49fi)  by  omitting  all  the  terms  involving  the  effects  of  the  loads  P, 
thus: 

Mmt  =  ~Xaty  cos  a  =Nmtv 

Nmt=     Xatcos((f>-a) 
Tmt=  -Xat  sin  (<£  -a) 

Abutment  displacements.  When  the  abutments  undergo  displacements  Ar  in  the 
directions  of  the  reaction  forces  R,  then,  for  the  case  of  external  redundancy  where 
da=0,  Eq.  (8D)  gives  for  P=0, 


ART.  49  SPECIAL  APPLICATIONS  OF  INFLUENCE  LINES  157 

where   Ra  has  the   several   values   Aa=-l-sina,   Ba  =  \-  sin  a,  and   Xa  =  l,   while   Jr 
represents  the  vertical  displacements  A  A,  AB  and  a  change  M  in  the  length  of  span  I. 
Hence  the  redundant  Xar,  due  only  to  abutment  displacements,  is 

S/g0Jr  _     Aa4 

Xar  —  ix 


or 

JB  sin  a  —  A  A  sin  a  +  M  sec  a. 

•X-ar  =  5~~'  '   "    "• 


When  AA=AB,  which  is  usually  the  case,  then 

.     (49L) 

showing  the  effect  due  to  an  elongation  M,  in  the  span  I,  to  be  precisely  the  same  in 
character  as  that  due  to  a  uniform  fall  in  temperaturegiven  by  Eq.  (49i). 

For  a  two-hinged  arch  with  a  tension  member  AB,  and  one  expansion  bearing,  the 
problem  becomes  one  involving  an  internal  redundant  member  Xa  and  Eq.  (42s)  is 
then  used  vvith  the  term  pa  in  the  denominator.  The  solution  would  otherwise  remain 

unchanged.  . 

Stresses  on  any  arch  section.  Neglecting  curvature,  which  is  always  permissible, 
the  stress  on  the  extreme  fiber  of  any  arch  section  mm'  is  given  by  Navier's  law  as 

N    My  (4j)M) 

J~F±  I  ' 

where  N  is  the  normal  thrust  on  the  section,  M=Nv  is  the  bending  moment  about  the 
gravity  axis  F  is  the  area  of  the  section,  y  is  the  distance  from  the  gravity  axis  to  the 
extreme  fiber  and  I=Fr*  is  the  moment  of  inertia  of  the  section  about  the  gravity  axis. 

Fig    49B    gives  a  graphic  representation  of  the  distribution  of  stress  on  any  arch 
section  mm'  in  accordance  with  Eq.  (49M).     The  center  of  gravity  is  at  c  and  the  uniform 
axial  stress  fm=N/F  is  the  stress  ordinate  at  c,  while  the  stresses  on  the  extreme  fibers 
/.  and  /,-,  are  applied  as   ordinates   giving  the  line  tt'   as   representing  the  uniform  , 
tribution  of  stress  over  the  section.  . 

The  point  e  is  the  kernel  point  for  the  extrados  while  * is  the  kernel  point  for  the  mtrados. 
These  points  are  determined  by  the  distances  k.=i*/e  and  k^fi/i,  where  r  is  the  radius 
of  gyration  of  the  section  and  e  and  i  are  the  respective  distances  to  the  extreme  f 
the  section  measured  from  the  gravity  axis  c. 

When  N  and  *  are  given,  the  line  tt'  is  easily  constructed  as  indicated  by  prolong- 
ing i?_to  6  to  find  t,  and  by  prolonging  ?ito  a  to  find  f.  The  stress  /.  is  then  the  inter- 
cept bn  while  the  stress  ft  is  the  intercept  an. 


158 


KINETIC  THEORY  OF   ENGINEERING  STRUCTURES  CHAP.  X 


Using  e  and  i  as  subscripts,  referring  respectively  to  extrados  and  intrados,  then 
for  M=Nv  and  I^Fr2,  Eq.  (49M)  gives 


and 


(4*0 


The  moments  about  the  kernel  points  e  and  i  are  evaluated  from  the  figure  as 

N(v+ke)=Me   and   N(v-kd=Mi  ........     (49o) 


FIG.  49B. 
For  r2/e=ke  and  r'2/i=k{,  Eqs.  (49N)  become 


-  _i  4.        =  _  -   _j? 

F\     +ke)  F\      ke      I  Fke 


F  \      ki/    ''F  \    ki   i     Fki 
Also  solving  Eqs.  (49o)  for  N  and  v,  then 

Me— Mi 

N=—, — -r—       and       v  = 


(49?) 


Me -Mi 


(49Q) 


Hence  Eqs.  (49p)  and  (49cj)  furnish  a  complete  solution  for  the  stresses  fe  and  /, 
on  the  extreme  fibers,  the  normal  thrust  N  and  its  distance  v  from  the  gravity  axis, 
for  any  unsymmetric  section  tt'  of  an  arch  ring,  in  terms  of  Met  Mi,  ke  and  &,. 


ART.  49 


SPECIAL  APPLICATIONS  OF  INFLUENCE   LINES 


159 


When  the  section  is  symmetric  about  a  horizontal  gravity  axis,  which  is  usually  the 
case,  then  these  equations  give,  for  e=i=D/2,  or  the  half  depth  between  extreme  fibers, 


_-  __      _ 
=  ,-i  _______ 

Me  Mi 

=s~  and        i== 


Me-Mi 

1.  T      "~       ,  Ctllvl 


i)     Me+Mi 


(49R) 


2k  Me -Mi  2N 

For  a  rectangular  section  of  unit  thickness  and  of  depth  D,  making  e=i=D/2. 

fc_»*  x>    F==1.D}    /=i^-3 

Me+Mi 


N=- 


and       ?'=: 


2N 


(49s) 


In  all  the  above  formulae  Me  and  Mi  have  opposite  signs  when  N  acts  between  the 
two  kernel  points  e  and  i.  When  v>ki,  both  moments  are  negative  and  when  v  is  negative 
and  larger  than  ke,  both  moments  are  positive.  The  figure  shows  v  to  be  positive 
when  measured  from  the  gravity  axis  at  c  toward  the  extrados.  The  stresses  fe  and 
fi  take  their  signs  from  Me  and  Mi  respectively,  and  compression  is  regarded  as  a  negative 
stress. 

The  stresses  on  any  arch  section  mm'  may,  therefore,  be  found  for  any  simultaneous 
position  of  a  moving  train  of  loads  when  the  influence  areas  for  Met  Mt  and  Tm  have 
been  drawn. 

The  resultant  polygon  for  any  simultaneous  case  of  loading  may  also  be  drawn  by 
plotting  the  offsets  v  from  the  gravity  axis  of  each  section  examined,  observing  that 
+v  is  toward  the  extrados  and  —  v  is  toward  the  intrados. 

The  method  of  combining  the  Xa  influence  line  with  ordinary  moment  and  shear 
influence  lines  to  obtain  the  Me,  Mi  and  Tm  influence  areas  for  any  section  mm'  will 
now  be  illustrated. 

Kernel  moment  influence  areas.  The  moment  equations  for  the  kernel  points  accord- 
ing to  Eq.  (49B)  are 


,_    \Moe     „!  [     Moe  1  , 

=Mae\  -a-, Xn  \=ye  cos  a  \- Xa\ 

lMae         J  [j/ecos  a 

' 


(49r) 


. 

The  intercept  of  the  Moe  influence  line  on  the  vertical  through  A  is  simply  the 
ordinate  xe  of  the  point  e.     Similarly  for  the  Moi  influence  line  this  intercept  is  the  ordinate 


160  KINETIC  THEORY  OF    ENGINEERING   STRUCTURES  CHAP.  3 

Xi  of  the  point  i.  Hence  the  positive  intercepts  on  the  vertical  through  A,  of  th« 
Moe/ye  cos  a  and  of  the  M^/yi  cos  a  influence  lines,  are  easily  computed  when  the  coordi- 
nates of  the  kernel  points  e  and  i  are  given,  provided  the  Xa  influence  line  was  drawn 
for  a  factor  /*  =  !. 

The  coordinates  of  the  kernel  points  are  derived  from  the  coordinates  (xmym)  of  the 
axial  point  for  the  same  section,  when  the  kernel  distances  k  and  the  angle  0  are  known 
for  that  section.  From  Fig.  49s, 

, 
xe  =  xm+kesm<f>,         ye  =  ym-kecos(j>  } 

I, V49u) 

X{  =xm  -ki  sin  0,         yi  =ym  +k{  cos  ^  J 

which  apply  to  all  points  from  A  to  the  crown,  and  from  there  to  the  abutment  B  the 
signs  of  the  last  terms  are  reversed.  By  using  the  I—  x  ordinates  the  intercepts  on 
the  vertical  through  B  are  found. 

One  intercept  only  need  be  computed  as  the  two  limiting  rays  of  the  M0/y  cos  a. 
line  must  intersect  on  the  vertical  through  the  center  of  moments.  See  Fig.  49c. 

By  constructing  the  Xa  influence  area  for  //  =  !,  which  may  always  be  done  by  making 
the  pole  distance  H  =  2wy  cos  a,  and  since  both  the  Xa  and  the  M0/y  cos  a  lines  are  pos- 
itive, then  by  applying  them  below  the  closing  line  A'W,  the  area  included  between  the 
two  lines  will  represent  the  Mm  influence  area.  The  portion  below  the  Xa  line  will 
be  the  positive  area  because  Xa  is  subtractive  in  Eqs.  (49-r).  The  entire  Mm  influence 
area  has  a  factor  p.  =y  cos  a  =Ma. 

The  Tm  influence  area  for  tangential  force  on  any  section  mm'.  From  Eqs.  (49u) 
the  following  equation  for  Tm  is  obtained. 


From  this  the  end  ordinate  at  A,  for  the  T0/TaYme,  becomes  1-  cos  <£^  sin  (0— a) 
because  the  end  ordinate  for  the  Q0  line  is  unity.  In  this  case  the  end  ordinate  at  B  is 
numerically  the  same  but  negative.  For  axial  points  to  the  right  of  the  crown  the  end 
ordinate  at  B  is  positive  and  the  one  at  A  is  negative.  Other  details  are  illustrated 
in  connection  with  the  example.  The  final  Tm  influence  area  has  a  factor  /i=sin  (<£-a). 

Example.  A  two-hinged  arched  rib,  modeled  after  the  Chagrin  River  Bridge  near 
Bentleyville,  O.,  was  selected  to  illustrate  the  application  of  the  previous  theory. 
The  structure  was  made  unsymmetric  by  shortening  the  span  at  the  right-hand  end  as 
shown  in  Figs.  49c.  The  clear  span  thus  became  164  ft.  and  the  rise  27.89  ft.  This 
bridge  was  designed  to  carry  a  live  load  of  24  kips  per  truss  per  panel.  The  arch  section 
is  composed  of  a  f-inch  web  plate  and  4-6"  X6"  X 11/16"  angles,  with  2-14"  X7/16"  flange 
plates  on  each  flange.  Besides  these  there  is  one  14"  X|"  plate  on  each  flange  extend- 
ing from  sections  2  to  5  and  7  to  10.  The  sections  are  all  symmetric  about  the  gravity 
axis  and  all  general  dimensions  are  given  on  the  drawing  and  in  Table  49A. 

The  bridge  consists  of  two  steel  arched  ribs  27  ft.  between  centers  and  carrying  a 
total  dead  load  of  1,058,000  pounds.  The  following  values  are  assumed  as  a  basis 


ART.  49 


SPECIAL  APPLICATIONS    OF  INFLUENCE  LINES 


161 


X  ordinores  are  ?.a  times 
I  toicole  of  Ittjfeth  ,. 


Me  INFLUENCE  AREA.  >\=ygCoiqC-18.-»l 


B' 


FIG.  49c. 


162 


KINETIC   THEORY   OF    ENGINEERING    STRUCTURES 


CHAP.  X 


for  the  analysis:     #=29,000  kips  per  sq.in.  =4,176,000  kips  per  sq.ft.;     e -0.0000065 
per  1°F.;    andZ  =  ±75°F. 

The  first  step  in  the  analysis  is  to  compute  the  Ew  loads  from  Eq.  (49r)  using  the 
tabulated  dimensions  in  Table  49A.  In  the  same  table  the  values  Ewy  cos  a  are  computed, 
giving  the  required  pole  distance  H=E2wycosa  for  constructing  the  Xa  influence  line 
in  accordance  with  Eq.  (49n). 

TABLE  49A 
Xa  INFLUENCE  LINE 


Sec- 
tion. 

Coordinates  of 
Gravity  Axis. 

<*> 

l 

/ 

in.4 

/ 

ft.« 

6  = 
1 

Jx 

ft. 

0J.r  = 
Ax 

Ew  = 
Axy  cos  a 

Ewy  cos  a 

X 

ft. 

y 
ft. 

cos  <t> 

cos  <f> 

ft, 

1  COS  (j> 

ft. 

I  COS  (j> 

ft. 

0 

0 

0 

36°  15' 

1.240 

74,180 

3.578 

0.3466 

9.4 
15 

15 
15 
15 

15 
15 

15 
15 
15 

15 

4.6 

0.652 

0 

0 

1 

9.4 

6.01 

32    15 

1.182 

74,180 

3.578 

0.3260 

4.954 

29.8 

179 

2 

24.4 

14.12 

26   00 

1.113 

64,710 
76,530 

3.121 
3.691 

0.3567 
0.3017 

4.939 

69.7 

984 

3 

39.4 

20.30 

20  00 

1.064 

66,170 

3.191 

0.3332 

5.014 

101.8 

2065 

4 

54.4 

24.73 

14    15 

1.032 

56,750 

2.737 

0.3770 

5.673 

140.3 

3468 

5 

69.4 

27.24 

7   30 

1.009 

48,160 
40,420 

2.323 
1.949 

0.4352 
0.5176 

7.198 

196.0 

5339 

6 

84.4 

27.89 

0  00 

1.000 

33,590 

1.620 

0.6173 

9.011 

251.3 

7007 

7 

99.4 

26.77 

7   30 

1.009 

40,420 
48,160 

1.949 
2.323 

0.5176 
0.4352 

7.198 

192.6 

5156 

8 

114.4 

23.58 

14    15 

1.032 

56,750 

2.737 

0.3770 

5.673 

133.7 

3153 

9 

129.4 

18.58 

20  00 

1.064 

66,170 

3.191 

0.3332 

5.014 

93.1 

1730 

10 

144.4 

11.82 

26  00 

1.113 

76.530 
64,710 

3.691 
3.121 

0.3017 
0.3567 

4.939 

58.4 

690 

11 

159.4 

3.14 

32    15 

1.182 

74,180 

3.578 

0.3260 

3.983 

12.5 

39 

12 

164.0 

0 

34   20 

1.211 

0.3260 

0.0 

0 

0 

EZwycos  a  =  29,810 

NOTE.    The  x  abscissae  are  measured  horizontally  from  A. 
The  y  ordinates  are  measured  vertically  from  AB. 
The  Jj  are  measured  horizontally  between  sections. 
a=l°  06',  cos  a  =  0.9998,  sin  a  =  0.0192. 

The  products  QAx  in  Table  49A  are  summed  by  the  prism oidal  formula  treating 
Jx  as  the  length  of  a  prismoid  whose  middle  area  is  6  and  whose  end  areas  are  the  means 
of  the  successive  tabulated  values  of  6.  This  is  necessary  because  6  is  a  variable  quantity. 


AHT.  19  SPECIAL  APPLICATIONS  OF  INFLUENCE    LINES  163 

Calling  0o-i  the  mean  between  00  and  0i,  and  0]-2  the  mean   between  6\  and  02 
etc.,  then  the  values  6 Ax  become: 

0Qjx  =^[200  +00_!]  =^[0.6932  +0.3363]  =0.652; 
o  o 

0!  Jz  =  ^[0o-i  +40i  +0i_2]  =—[0.3363  + 1.3040  +0.3413]  =4.954. 

62Ax  =^[0i  -2  +202  +202'  +02-3]  =^?[0.3413  +0.7134  +0.6034  +0.3174]  =4.939. 


6zAx  =  ^[02-3  +403+03-4]  =y  [0.3 174  +  1.3328  +0.3551]  =5.014, 
etc.,  etc. 

After  collecting  the  tabulated  data  from  the  drawing,  Fig.  49c,  the  remaining 
computations  are  quite  simple  and  no  further  comment  is  necessary  here,  as  the  several 
operations  are  indicated  in  Table  49A. 

The  Xn  influence  line  is  now  drawn  by  combining  the  Ew  loads  into  a  force  polygon 
and  making  the  pole  distance  equal  to  E2wy  cos  a.  The  resulting  equilibrium  polygon 
represents  the  Xa  influence  line  with  ordinates  to  the  scale  of  lengths  and  /*  =  !,  in 
accordance  with  Eq.  (49n). 

In  the  drawing,  the  pole  distance  was  made  1/20  of  the  actual  length  so  that  the 
ordinates  ??„  are  twenty  times  too  large.  For  this  reason  a  special  scale  of  ordinates, 
twenty  times  the  scale  of  lengths,  was  constructed  and  used  for  all  influence  ordinates. 

The  scale  of  forces  for  the  force  polygon  may  be  any  convenient  scale,  so  long  as  the 
pole  distance  is  measured  with  the  same  scale  as  the  forces. 

The  influence  areas  for  the  kernel  moments  and  tangential  force  are  now  drawn  by  com- 
bining the  M0/Ma  lines  and  the  T0/Ta  line  each  with  the  Xa  line.  According  to  Eqs. 
(49x),  (49u),  (49v),  this  will  require  computing  the  kernel  distances  k,  and  the  coordi- 
nates of  the  kernel  points  and  finally  the  end  ordinates  at  A  of  all  influence  lines. 
The  end  ordinates  at  B  may  be  obtained  by  substituting  for  the  abscissa?  x  the  values 
I—  x.  These  computations  are  given  in  Table  49s,  which  is  self-explanatory. 

In  Fig.  49c  the  influence  areas  Me  and  Mi  are  constructed  for  section  3.     The  Xa 
polygon  is  copied  from  the  original  one  by  transferring  down  the  several  j)a  ordinates 
from  a  horizontal  base'  ~ArW.     The  M0/Ma  lines  are  constructed  from  the  end  ordinates 
A'A"=xc/yecosa    and   AfA7r'=Xi/yicosa,   using   the   coordinates   for   the    two   kernel, 
points  as  given  in  Table  49u.     The  lines  WA"  and  B'A'"  are  thus  determined. 

The  kernel  points  e  and  i  are  the  centers  of  moments  and  have  the  abscissae  xe  and 
xi}  as  given  in  the  table,  and  from  these  the  points  e'  and  i'  are  located. 

The  two  moment  influence  lines  are  completed  by  drawing  the  lines  A'e'  and  A'i'. 

The  Moe/Mae  influence  line  is  thus  found  to  be  the  broken  line  A'e'B'  and  the  shaded  area 

included  between  it  and  the  Xa  line   is  the  Me  influence  area  for  section  3.     The  fac- 

tor  for  the   ordinates   i)e,  when   measured  to  the    same  scale  as  the   rja  ordinates,  is 

i  =     cos  a  =18.41. 


164 


KINETIC  THEORY   OF  ENGINEERING  STRUCTURES 


CHAP.  X 


TABLE  49n 
KERNEL   POINTS  AND  END   ORDINATES   FOR   Alme,  Mmi,  AND  Tm   AREAS 


Sec- 
tion. 

F 

Sq.  in. 

D 

2 

in. 

i       2I 

* 

k  sin  <£ 
ft. 

k  cos  <j> 
ft. 

Coordinates  of  kernel  points, 
Eqs.  (49u). 

End  Ordinatea. 

\-2FD 

Eq.  (49R) 

ft. 

xe 
ft, 

Ve 
ft. 

xi 
ft. 

»* 

ft. 

xe. 

xi 

COS  <ji 

ye  cos  a 

y^  COS  a 

sin  (<t>—  a) 

0 
1 
2 

3 

4 
5 

6 

7 

8 
9 
10 

11 
12 

81.6 
79.0 
89.5 
87.8 
86.3 
85.7 
75.2 
73.6 
75.2 
85.7 
86.3 
87.8 
89.5 
79.0 
81.6 

36.0 
35.5 
33.4 
33.8 
31.6 
29.5 
27.4 
27.0 
24.9 
27.0 
27.4 
29.5 
31.6 
33.8 
33.4 
35.5 
35.7 

2.13 
2.04 
2.11 
1.99 
1.86 
1.71 
1.64 
1.53 
1.64 
1.71 
1.86 
1.99 
2.11 
2.04 
2.13 

36°  15 
32    15 

26    00 
20    00 
14    15 

7    30 
0   00 

7    30 
14    15 
20    00 

26    00 
32    15 
34    20 

1.14 

0.91 

0.68 
0.46 

0.23 
0.00 

0.23 
0.46 
0.68 

0.91 
1.14 

1.80 

1.86 
1.87 
1.79 

1.66 
1.53 

1.66 
1.79 
1.87 

1.86 
1.80 

10.54 

25.31 

40.08 
54.86 

69.63 
84.40 

99.17 
113.94 
128  .  72 

143.49 
158.26 

4.21 

12.26 
18.43 
22.94 

25.58 
26.36 

25.11 
21.79 
16.71 

9.96 
1.34 

8.26 

23.49 
38.72 
53.94 

69.17 

84.40 

99.63 
114.86 
130.08 

145.31 
160.54 

7.81 

15.98 
22.17 
26.52 

28.90 
29.42 

28.43 
25.37 
20.45 

13.68 
4.94 

2.504 

2.064 
2  .  174f 

1.058 

1.470 
1.747t 

1.633 

2.135 

2  .  895f 
4.307 

8.884 
55.555 

8.884* 
4.307 
2.895 

2.135 
1.633 

3.202 
2  .  582* 

2  .  869 
2.265* 

4.284 

0.700 

NOTE.     D  is  measured  between  extreme  fibers.     For  all  dimensions  in  inches,  k  is  in  inches. 
The  coordinates  of  the  axial  points  are  given  in  Table  49A. 

*  For  sections  7  to  12  the  ordinates  are  for  the  B  end. 
t  Values  used  in  Fig.  49c. 


Similarly  the  dotted  line  A'i'B'  represents  the  Moi/Mai  influence  line  and  the  area 
included  between  it  and  the  Xa  line  represents  the  Mi  influence  area  with  factor 
Pi=yi  cos  a  =22.13  for  section  3. 

Hence  for  any  simultaneous  position  of  moving  loads  the  two  kernel  moments  for 
section  3  are  represented  by  the  expressions 


and 


(49w) 


where  the  subscripts  refer  to  the  kernel  points. 

The  Tm  influence  area  for  section  3,  is  found  by  laying  off  the  end  ordinate  A'  A"  = 
cos  <£/sin  ((f>  —a)  =2.895  and  drawing  the  line  A"B'.  The  negative  end  ray  A'  3'  is  parallel 
to  A"B',  and  the  T0/Ta  line  is  thus  the  broken  line  A'3'3"B'  and  the  shaded  area  is  the 
T3  influence  area  with  a  factor  /*=sin  (<£  —  a). 

It  will  be  seen  that  for  sections  near  the  crown  the  end  ordinate  at  A  becomes  very 
large  and  when  <£=«,  this  ordinate  approaches  infinity,  while  «  becomes  zero.  Hence 
there  will  always  be  several  sections  near  the  crown  for  which  the  Tm  influence  area 
must  be  found  by  making  n  =  l  and  reducing  the  Xa  line  accordingly.  This  is  done 
by  using  the  original  form  of  Eq.  (49s)  which  is 

Tm  =Q0  cos  <£  —  Xa  sin  (<j>  —a). 


ART.  49  SPECIAL  APPLICATIONS  OF  INFLUENCE  LINES  165 

For  the  A  ordinate  Q0  =  l,  hence  this  ordinate  is  simply  cos  <£  and  by  reducing  all 
the  T)U  ordinates  by  multiplying  them  by  sin  (<£-«),  the  .T5  area  is  found  with  a  factor 

1=1. 

The  rja  ordinates  are  easily  reduced  by  graphics,  laying  off  the  line  B'z  such  that  its 
deviation  from  the  vertical  is  sin  (0  -a)  in  a  distance  unity.  The  rja  ordinates  pro- 
jected over  horizontally  are  then  reduced  to  the  small  horizontal  intercepts  between 
Wz  and  the  vertical. 

The  T5  influence  area  is  thus  constructed  and  is  represented  by  the  shaded  area 
included  between  the  broken  line  A'5'5"B'  and  the  reduced  Xa  line.  The  factor  p  =  l. 

The  tangential  force  for  section  3,  for  any  train  of  moving  loads  is  expressed  by 


(49x) 


and  the  tangential  stress  by  TZ/F3,  where  rjt  is  any  ordinate  to  the  T3  area  under  some 
particular  load  P. 

The   stress   due   to   a  unifo 
(49  J)  whence  for  t  =  ±  75°  F., 


particular  load  P. 

The   stress   due   to   a  uniform  change   in   temperature   is    found    from   Eqs.    (49i)    and 


•l-.a*_  etf  0.000488  1 

Xat  —     5         —  it  ?,       „„ 


daa  COS  Oi  daa   COS  a 

Eq  (49o)  gives  daa  =  Hwycosa  and  Table  49A  furnishes  EZwij  cos  <£  =29810,  hence 
L=29810/JS  ft.  for  Xa  =  l  kip.  Making  £=4,176,000  kips  per  sq.ft.,  Z  =  164  ft.  and  cos 
aa=0.999S,  then  0^-0.00714  ft.  and  Xat  =  ±0^mfxo.9998=  ±n-21  kiPs- 

*    Eqs.  (49K) ,  will  furnish  the  means  of  finding  the  values  Mmt,  Nmt,  Tmt,  and  v=Mmt/Nmt 
and  with  these^and  Eqs.  (49?)  the  temperature  stresses  fe  and  ft  may  be  computed  for 

any  section  mm'. 

Stress  due  to  abutment  displacements.  Eq.  (49L)  furnishes  the  haunch  thrust  Xar 
for  any  change  Al  in  the  length  of  span.  The  quantity  Al  must  be  estimated  from  the 
elastic  properties  of  the  abutments  and  is  always  more  or  less  problematic,  though  it 
is  well  to  investigate  the  probable  stress  which  might  be  created  by  such  a  displacement. 

In  the  present  example  it  is  assumed  that  for  the  maximum  case  of  loading  the 
haunches  will  spread  an  amount  JZ=0.03  ft,  then  for  daa  =0.0071<  ft,  Xa-  kip, 
and  cos  a  =0.9998,  Eq.  (49L)  gives 

y  &  °-03 =  -4.20  kips. 

ar  =  ~<LTc^         0.00714X0.9998 

The  stresses  /,  and  /,-  may  then  be  found  in  precisely  the  same  manner  as  was 
just  described  for  the  case  of  temperature  stresses. 

The  stresses  due  to  temperature  and  abutment  displacements  should  be  separately 
investigated  for  several  typical  sections,  especially  the  crown  section,  so  as  to 
thldesigner  to  judge  of  the  relative  importance  which  these  may  have  in  comparison 
with  the  combined  dead  and  live  load  stresses. 

The  live  load  stresses  for  section  3  will  now  be  found  for  a  single jloac  P  acting  at  3, 
tion  4,  merely  to  illustrate  the  use  of  the  influence  areas  shown  in  Pig.  4 


166  KINETIC   THEORY   OF   ENGINEERING  STRUCTURES  CHAP  X 

Me3=Me4P  =  18.41  X0.45P  =  8.285P. 
M    =i*P  =22.13  X0.17P  =3.762P. 


=  -6.84P  per  sq.ft.  =  -0.0474P  per  sq.in. 


M,-       3.762P 


=  LOOP;       T3  =pti)t*P  =0.324  X0.94P  =0.305P. 


The  resultant  KA,  of  ^1  and  JTa,  is  found  graphically  to  be  1.19  P,  which,  for  a  single 
load  to  the  right  of  section  3,  should  check  R3,  as  it  does. 

ART.  50.     TWO-HINGED   SPANDREL   BRACED   ARCH  OR  FRAMED  ARCH 

The  analysis  of  the  two-hinged  frame  arch  is  accomplished  in  the  same  general  manner 
followed  in  the  previous  article,  where  the  corresponding  solid  web  arch  was  treated, 
only  that  the  frame  is  usually  simpler. 

The  general  Eqs.  (49s)  for  the  reactions  A,  B,  and-  moment  Mm  apply  equally  to  an 
unsymmetric  framed  arch,  and  the  influence  line  for  the  redundant  haunch  thrust  Xa 
is  determined  precisely  as  in  the  previous  problem  except  that  the  w  loads  are  found  by 
the  method  given  in  Art.  36. 

The  stress  in  any  member  then  becomes 

S=S0—SaXa=Sa\-^-—Xa\,     ........     (50A> 

according  to  Eq.  (45E)  for  one  external  redundant  condition.  This  equation  is  later 
employed  to  construct  the  stress  influence  lines  for  the  members. 

The  Xu  influence  line  for  the  haunch  thrust  is  represented  by  Eq.  (49n)  as 

Z*  '  Oma  _          •!•  '  O-ma  f~p.    \ 

a  ™=      K          """v*  —  'Ja?         ........       (OUB) 

daa  2jWy  COS  a 

where  d^a  is  the  vertical  deflection  ordinate  of  any  point  m  of  a  deflection  polygon,  drawn 
for  the  loaded  chord  and  for  the  conventional  loading  Xa  =  l  applied  to  the  principal 
system;  and  daa  =  '£wy  cos  a  is  the  change  in  the  span  AB,  due  to  Xa  =  l  acting  on  the 
principal  system. 

According  to  Art.  40,  the  dma  deflection  polygon  is  the  equilibrium  polygon  drawn  for  a 
set  of  elastic  loads  w,  with  a  pole  distance  H  =  l,  according  to  the  method  given  in  Art.  36c. 


ART.  50  SPECIAL  APPLICATIONS  OF  INFLUENCE  LINES  167 

The  elastic  loads  w  are  functions  of  the  changes  in  the  lengths  of  all  the  members 
of  the  frame  as  given  by  Eqs.  (36fi)  and  (36c).  These  are  algebraically  summed  for  all 
the  panel  points  to  obtain  the  total  loads  w.  Thus  the  elastic  loads  for  the  chords  alone 
are  wc  =  Al/r  and  each  web  member  contributes  two  elastic  loads  wu  =  M/ru  and  wn  =  M/rn  , 
acting  at  the  two  adjacent  panel  points  u  and  n  of  the  loaded  chord.  Hence  when  the 
loads  P  are  to  be  applied  to  the  top  chord,  then  the  wu  and  wn  elastic  loads  are  com- 
puted for  the  top  chord  panel  points.  The  lever  arms  r  for  the  chords  are  measured 
as  shown  in  Fig.  36A  and  the  lever  arms  ru  and  rn  for  the  web  members  are  measured  as 
explained  in  Figs.  36s  and  36r.  The  details  of  the  computation  of  the  w  loads  are  illus- 
trated in  Table  oOu,  and  Fig.  50A,  in  connection  with  a  complete  example. 

For  any  braced  arch  with  parallel  chords,  the  w  loads  for  the  web  members  may 
always  be  neglected. 

The  displacement  daa  =  'Swy  cos  a  is  computed  for  the  same  w  loads  by  taking  the 
sum  of  their  moments  about  the  line  A  B  joining  the  haunches.  The  lever  arms  for  an 
unsymmetric  span  thus  become  the  vertical  ordinates  of  the  respective  pin  points 
times  cos  a.  See  also  Table  50A. 

The  Xa  influence  line  thus  becomes  the  equilibrium  polygon  drawn  for  the  elastic 
loads  w  with  a  pole  distance  oaa  =  ^wy  cos  a. 

Still  another  method  of  finding  this  influence  line  consists  in  drawing  the  deflection 
polygon  for  the  loaded  chord  by  means  of  a  Williot-Mohr  displacement  diagram,  which 
also  furnishes  the  value  d^  from  which  the  influence  line  ordinates  rja  may  be  computed 
and  plotted.  This  solution  is  illustrated  in  Fig.  50B. 

Since  the  modulus  E  does  not  affect  the  Xa  influence  line  it  is  more  convenient  to 
comput-e  all  displacements  and  the  w  loads  E  times  too  large,  thus  avoiding  the  small 
quantities  resulting  in  many  decimals. 

Temperature  Stress-  Calling  Alt  the  change  in  the  length  of  any  member  due  to  any 
temperature  effect,  then  from  Mohr's  work  Eq.  (5n)  the  change  dat  in  the  length  of 
the  span  AB  becomes 

l.dat=ZSa4lt=2Sadl  ..........     (50c) 

The  redundant  thrust  Xat,  due  to  any  temperature  effect,  is  found  from  Eq.  (44c)  as 


.at_a_  g 

at~~d^~    daa     'EZwycosa 

For  a  uniform  change  in  temperature  of  ±t°,  above  or  below  a  certain  normal,  the 
change  o^  in  the  length  of  the  span  'AB,  is  found  from  Eq.  (49j)  as 

dAB 


cos  a 


In  any  case  the  stresses  in  the  members  are  best  found  from  St  =  ^SaXat  for  ±Xat,  or 
from  a  Maxwell  diagram  drawn  for  the  external  loading  Xat,  acting  on  the  principal  system. 

For  a  uniform  change  of  t°  from  the  normal  temperature,  the  resulting  stresses 
become  St  =  ^SaXat.  The  effect  on  the  final  stresses  ±S  for  full  loading,  will  then 
be  additive  or  ±Smax  =  ±(S  +St.) 


168  KINETIC   THEORY   OF    ENGINEERING   STRUCTURES  CHAP.  X 

Abutment  displacements  will  affect  the  stresses  in  the  members  by  producing  a  redundant 
thrust  —  Xar  as  found  from  Eq.  (49L)  when  JZ  is  an  increase  in  the  length  of  span  AB. 
The  stresses  Sr  would  be  found  from  a  Maxwell  diagram  drawn  for  —  Xar  acting  on  the 
principal  system.  They  could  also  be  found  as  Sr  =  =F*Sa  Xar,  for  ±  Xar,  since  *Sa  is  the  stress 
produced  in  any  member  S  by  Xa  =  1  acting  outward. 

The  stress  influence  area  for  any  member  $  is  derived  from  the  previous  Eq.  (50 A) 
from  which  any  stress  influence  ordinate  TJ  becomes 


s=sa  ^-xa^=y).   .:.•;  .  . (SOP) 

These  influence  areas  are  alike  in  principle  for  all  members,  whether  chords  or  web 
members,  and  represent  the  area  inclosed  between  the  Xa  influence  line  and  the  S0/Sa 
influence  line,  times  a  factor  Sa. 

The  lSo/Sa  influence  line  is  the  ordinary  stress  influence  line  S0  for  any  determinate 
frame,  nultiplied  by  the  factor  l/Sa.  Hence  the  end  ordinates  for  the  S0  line  may  be 
found  in  the  usual  way.  Thus  the  stress  SA  in  the  member  S  due  to  the  upward  reaction 
.4=1  is  the  end  ordinate  of  the  S0  line  at  the  point  A,  and  this  ordinate  divided  by 
Sa  becomes  the  required  end  ordinate  rjA=SA/Sa  of  the  S0/Sa  line  at  A.  Similarly  the 
end  ordinate  of  the  S0/Sa  line  at  B  is  -fjB=SB/Sa,  where  SB  is  the  stress  in  the  member 
due  to  the  upward  reaction  B  =  \  acting  on  the  principal  system. 

Hence  if  the  stresses  SA,  SB  and  Sa,  for  all  the  members  of  the  principal  system,  are 
found  either  by  computation  or  from  three  Maxwell  diagrams,  then  all  the  data  for  the 
several  stress  influence  lines  are  at  hand  provided  the  Xa  influence  line  is  known. 

In  drawing  the  S0/Sa  influence  lines  the  following  points  should  be  observed:  1, 
That  the  end  bounding  lines  must  always  intersect  in  a  point  i'  on  the  vertical  through 
the  center  of  moments  i  of  the  particular  member  treated.  2,  that  this  S0/Sa  line 
must  be  a  straight  line  between  adjacent  panel  points  of  the  loaded  chord.  3,  that  when 
one  of  the  end  ordinates  is  too  large  to  be  conveniently  measured,  then  half  this  ordinate 
may  be  laid  off  at  the  center  of  the  span.  This  is  frequent!}'  done  as  in  the  example 
which  follows,  see  Figs.  oOc. 

The  signs  of  the  end  ordinates  and  of  the  factors  fJi=Sa  all  follow  from  the  signs  of 
the  stresses  SA,  SB  and  Sa. 

Deflection  of  any  point  m.  Applying  Mohr's  work  equation  (Gfi)  the  deflection 
d,n  of  any  point  m,  becomes 


(600) 

SI 


wherein  Si  is  the  stress  in  any  member  due  to  a  load  P  =  l,  applied  at  the  point  m  in 
the  direction  of  the  desired  deflection ;  and  S  is  the  corresponding  stress  due  to  any  cause 
or  actual  condition  of  simultaneous  loading  for  which  the  deflection  is  sought,  includ- 
ing load  effects  as  well  as  temperature  changes  and  abutment  displacements.  The  sum- 
mation covers  all  the  members  of  the  principal  system. 


ART.  50 


SPECIAL  APPLICATIONS  OF  INFLUENCE  LINES 


169 


For  any  redundant  condition  Xa  the  stress  S  by  Eq.  (7 A)  becomes 


S=S0-SaXa+St,     . (50H) 

where  St  =  —SaXat  is  the  stress  due  to  temperature  effect. 

Example.  A  two-hinged,  riveted,  spandrel-braced  arch,  taken  from  a  thesis  by 
Mr.  A.  V.  Saph,  1901,  Cornell  University,  is  used  to  illustrate  the  above  method. 

The  arch  has  a  span  of  168.75  ft.  between  pin  supports,  a  rise  of  29.5  ft.,  and  weighs 
1,060 ,320  Ibs.  without  abutment  shoes,  making  a  uniform  dead  load  of  43.2  kips  per  truss 
per  panel.  The  top  chord  is  the  loaded  chord.  The  abutments  are  symmetric  and, 
therefore,  a  =0. 

Fig.  50A  shows  the  half  span  with  the  lengths  of  members  in  feet  below  the  lines, 
and  the  values  EM,  in  feet,  as  found  from  Table  50A,  above  the  lines.  The  various  lever 
arms  used  in  the  computation  of  the  stresses  SA,  SB  and  Sa,  and  of  the  w  loads,  are  also 
shown.  The  actual  values  Al  in  feet  =EM  +  29000,  because  the  areas  F  were  not  reduced 
to  square  feet.  See  also  the  example  in  Art.  52,  where  M  is  actual. 

TABLE  50A 
COMPUTATION  OF  EAl,  j)A,  TJB  AND  E2Sa4lt. 


Stresses  in  Kips. 

Unit 

End  Ordinates, 

Temperature  Effect. 

Area, 

..enscth 

Stress, 

Sal 

Oo 

*ESa4lt 

Member 

SA 

SB 

Sa 

F 

i 

Oa 

F 

*EMt 

for 

for 

for 

* 

SA 

SB 

-stlh 

A  =  l 

5=1 

a 

Sq.in. 

ft. 

Kips.  sq.  in. 

ft. 

/      Sa 

T>B     Sa 

ft. 

Kip.ft. 

Kip.  ft. 

uu 

-   0.320 

-   5.437 

-0.211 

19.80 

15.000 

-0.0107 

-0.1605 

1.517 

25.770 

197.9 

41.8 

u  c/ 

-    1.165 

-   6.902 

-0.697 

19.80 

15.000 

-0.0352 

-0.5280 

1.671 

9.917 

197.9 

137.9 

u.u, 

-   2  725 

-   8.953 

-1.457 

19.80 

15.000 

-0.0736 

-  1  .  1040 

1.870 

6.145 

197.9 

288.3 

-   5  588 

-11.755 

-2.649 

26.48 

15.000 

-0.1000 

-  1  .  5000 

2.109 

4.438 

197.9 

524.2 

ifu* 

-10.008 

-14.336 

-4.121 

38.11 

15.000 

-0.1081 

-1.6215 

2.429 

3.479 

97  9 

815.5 

u*u* 

-14.062 

-14.062 

-4.917 

38.11 

15.000 

-0.1290 

-1.9350 

2.859 

2.859 

197.9 

973.1 

T  V    " 

01  79 

+  1  085 

42  50 

11  228 

+  0.0255 

0.2863 

-0.159 

88.9 

96.5 

LI 

0  366 

6.231 

1.388 

40.00 

17.190 

0.0347 

0.5965 

0.263 

4.481 

136.1 

188.9 

LI* 

1  269 

7.516 

1.848 

37.50 

16.335 

0.0493 

0.8053 

0.687 

4.067 

129.4 

239.1 

L  L 

2  857 

9  .  386 

2.575 

37  .  50 

15.725 

0.0687 

1.0803 

1.109 

3.645 

124.5 

320.6 

5  685 

11.958 

3.712 

37.50 

15.258 

0.0990 

1.5105 

1.532 

3.221 

120.8 

448.4 

J*  L 

10  027 

14  .  363 

5.131 

42.50 

15.029 

0.1207 

1.8140 

1.954 

2.799 

119.0 

610.6 

T?  L 

0  917 

-0.605 

23  .  52 

35.942 

-0.0257 

-0.9237 

1.516 

379  .  5 

229.6 

U°T° 

1   179 

-   2.043 

-0.678 

19.80 

29.312 

-0.0342 

-1.0025 

1.739 

3.014 

309.5 

209.8 

r/1/"1 

—    1  502 

-    1.976 

-0.732 

14.70 

20.918 

-0.0498 

-1.0417 

2.052 

2.700 

220.9 

161.7 

U>L3 

-    1.857 
2  042 

-   1.817 
-   1  193 

-0.773 
-0.681 

11.76 
11.76 

14.450  -0.0657 
9.730-0.0579 

-0.9494 
-0.5634 

2.402 
2.999 

2.351 
1.752 

152.6 
102.7 

118.0 
70.0 

U  L 

-    1.622 

+  0.109 

-0.318 

11.76 

6.932 

-0.0270 

-0.1872 

5.101 

-0.343 

73.2 

23.3 

77  V5 

0  0 

0  0 

0.0 

11.76 

6.000 

0.0 

0.0 

0.0 

0.0 

63.4 

0.0 

rr  r  ° 

If)    «71 

13  34 

32  928 

0  .  0503 

1.6563 

1.515 

347.7 

233.3 

T7T 

.UI/ 
1  451 

2  514 

0.834 

13.34 

25.741 

0.0625 

1.6088 

1.740 

3.015 

271.8 

226.7 

1 

2.166 
3.413 
4.869 
4.367 

2.848 
3.339 
2.843 
-  0.294 

1.055 
1.421 
1.622 
0.857 

13.34 
13.34 
18.25 
15.84 

20.828 
17.871 
16  .  524 
16.156 

0.0791 
0.1065 
0.0889 
0.0541 

1.6475 
1.9033 
1.4690 
0.8640 

2.053 
2.402 
3.002 
5.096 

2.700 
2.350 
1.753 
-0.343 

220.0 
188.7 
174.4 
170.6 
Tot'k 

232.1 
268.1 
282.9 
146.2 
3293.4 

3593  .  2 

*  Where  £=29,000  kips  instead  of  29,000X144.                                 lE^SaMt-      299.8  kip.  ft. 

170 


KINETIC  THEORY   OF   ENGINEERING  STRUCTURES 


CHAP.  X 


U  ' 


The  f  i^ure]  obove-the  members|  represent  EA] 
other  dijnen&ions    are  lengths   in  ft. 


DEF-LEcflONS  ARE    \E    TIlllES  ACTUAL    IN  F*ET. 


E=  29OOto  KIPS  PER  SQ.IN.          S 
\  .XI 


"      ":  r" 


AKT.  50  SPECIAL  APPLICATIONS  OF  INFLUENCE  LINES  171 

Table  50A  gives  these  stresses,  in  kips,  as  found  by  computation,  also  the  areas  F 
of  the  members,  being  gross  for  compression  and  net  for  tension  members.  The  lengths 
I  of  the  members,  unit  stresses/  and  quantities  EAl  (—for  shortening)  are  also  included 
in  this  table,  and  finally  the  end  ordinates  of  the  S0/Sa  influence  lines  are  obtained. 

The  Xa  influence  line  is  now  found  by  two  different  methods  to  illustrate  the  applica- 
tions frequently  referred  to  elsewhere. 

The  first  method  is  by  constructing  a  deflection  polygon  of  the  top  chord  by  means 
of  a  Williot-Mohr  displacement  diagram,  Fig.  50B,  using  the  quantities  EM,  in  feet,  as 
the  changes  in  the  lengths  of  the  members  for  the  conventional  loading  Xa  =  l,  producing 
stresses  Sa.  The  second  method  is  by  finding  this  deflection  polygon  from  the  w  loads. 

The  first  method  requires  little  description  other  than  to  say  that  the  displacement 
diagram  Fig.  50B  is  drawn  for  displacements  EAl,  Table  50A,  on  the  assumption  that  the 
point  LQ  and  the  direction  of  the  member  L6U6  remain  fixed  and  since  the  span  is  symme- 
tric about  this  member  and  the  point  L6,  therefore  no  rotation  diagram  is  necessary. 
The  deflection  polygon  of  the  top  chord  is  then  found  by  projecting  the  vertical  deflec- 
tions of  the  top  chord  panel  points  onto  the  verticals  through  these  points,  furnishing 
the  polygon  ~ArTlT§y  with  a  closing  line  A'A",  horizontally  through  A'  '.  These  deflec- 
tions are  E  times  actual,  in  feet,  measured  to  the  scale  of  displacements. 

The  horizontal  displacement  \Edaa  between  A  and  L6  is  also  obtained  from  the  same 
diagram  as  the  horizontal  distance  between  A'  and  L'6. 

The  vertical  ordinates  of  the  deflection  polygon  are  values  of  dma,  and  hence  the 
Xa  influence  line  ordinates  ya  are  found  by  dividing  the  several  deflection  ordinates 
Eoma  by  the  constant  Edm  according  to  Eq.  (50s),  giving 


This  shows  that  the  factor  E  and  the  scale  of  the  ordinates  do  not  affect  Xa.  The 
ordinates  r]f!,  for  all  panel  points,  are  plotted  to  any  convenient  scale  to  obtain  the  Xa 
influence  line. 

Second  method.  The  same  displacements  EM,  in  feet,  are  here  employed  to  com- 
pute the  elastic  loads  Ew,  using  the  lever  arms,  also  in  feet,  as  given  on  Fig.  50A.  See 
Table  50s. 

In  the  present  example  all  the  members  are  included  and  the  table  indicates  exactly 
how  much  each  member  contributes  to  the  several  total  Ew  loads.  The  method  of 
Art.  35  is  rigidly  followed.  By  using  displacements  EM  which  are  E  times  too  large, 
the  w  loads  are  also  multiplied  by  E. 

Each  chord  member  furnishes  one  w  load  which  acts  at  the  center  of  moments 
for  that  chord  while  each  web  member  contributes  two  loads  wu  and  wn  acting  at  the  two 
adjacent  panel  points  u  and  n  of  the  loaded  chord  or  the  chord  for  which  the  deflection 
polygon  is  to  be  drawn. 

"  The  wc  loads  resulting  from  the  chord  members  are  always  positive  as  found  by 

Eq.  (36s),  thus: 

M  „        EM  ro  , 

uv  =  —       or       Ewc=—        .........    (DUK; 


172  KINETIC  THEORY  OF  ENGINEERING  STRUCTURES         CHAP,  x 

The  negative  and  positive  loads  for  a  web  member  are 

Al  Al  EM  EAl 

wu=-       and      wn  =  —       or      Ewu=—         and      Ewn=  --     .     .     (50L) 

TU  rn  TU  Tn 

as  given  by  Eq.  (36o). 

The  lever  arm  r  for  a  chord,  is  the  distance  from  the  center  of  the  moments  for 
such  chord  and  perpendicular  to  the  chord.  The  lever  arms  ru  and  rn  are  perpendicular 
distances  onto  a  web  member  as  described  in  Figs.  36s,  and  36F.  They  may  be  identified 
by  comparing  the  values  in  Table  50s  with  the  dimensioned  lever  arms  on  Fig.  50A. 

The  rule  for  the  signs  of  the  two  loads  for  any  web  member  was  stated  in  Art.  36, 
and  is  as  follows:  Calling  all  top  chord  members  negative  and  all  bottom  chord  members 
positive,  and  giving  the  proper  sign  to  Al  for  the  web  member  in  question,  then  the  positive 
w  load  is  found  on  that  side  of  the  panel  where  the  sign  of  Al  coincides  mth  the  sign  of  the 
adjacent  chord. 

For  the  panel  point  f/0  the  total  load  Ew0  (Table  O()B)  is  made  up  of  the  positive 
load  produced  by  the  chord  L0Li  and  one  of  the  loads  from  each  of  the  web  members 
U0Li  and  U\L\,  the  signs  of  which  are  negative  according  to  the  above  rule. 

The  w  load  for  point  A  produces  zero  effect  on  the  deflection  and  need  not  be 
considered. 

For  the  panel  point  Ul  the  total  load  Ew\  becomes 

ALL      JUpLi     AVjLz  .  JE/nZq     J£72L2 

"  ""'  ~~    =auby 


25.7         13.4        17.15  "'    11.7 

and  similarly  for  the  other  panel  points  of  the  loaded  (top)  chord. 

The  bottom  chord  points  receive  only  the  loads  produced  by  the  top  chord  members 
except  at  LI  where  the  member  U0L0,  being  the  end  post,  contributes  a  load  Jt/0L0/10'.2. 

The  total  Ew  loads  acting  in  the  same  vertical  are  then  summed  for  the  seven  panel 
points  of  the  half  span  and  used  in  constructing  the  deflection  polygon  with  a  pole  dis- 
tance H  =E2yw. 

The  y  ordinates  are  measured  vertically  from  the  line  A~B  to  the  points  of  application 
of  the  respective  Ew  loads,  distinguishing  between  the  loads  acting  at  the  top  and 
bottom  panel  points. 

In  order  that  the  Xa  influence  line  ordinates  may  appear  to  a  scale  twenty  times 
as  large  as  the  scale  of  lengths,  the  pole  distance  was  made  equal  to  E2yw/20,  see  Fig. 
50c.  Note  the  agreement  of  the  ordinates  here  found  with  those  obtained  in  Fig.  50fi. 

Stress  influence  areas.  The  end  ordinates  i)A  and  r)B  of  the  S^/Sa  lines  are  com- 
puted in  Table  50A  and  these  serve  to  construct  all  stress  influence  areas  for  the  several 
members. 

Fig.  50c  shows  stress  influence  areas  for  six  typical  members,  and  each  one  has  a 
factor  Sa  as  per  Eq.  (50A).  The  end  ordinates  are  applied  down  from  the  closing  line 
when  positive  and  up  when  negative.  The  signs  of  the  influence  areas  are  uniformly  + 
for  areas  below  the  .Ya  influence  line  and  the  factor  p  takes  the  sign  of  Sa. 


SPECIAL  APPLICATIONS  OF  INFLUENCE  LINES 


173 


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KINETIC  THEORY  OF   ENGINEERING  STRUCTURES  CHAP.  X 


FIG.  50c. 


ART.  50  SPECIAL  APPLICATIONS  OF  INFLUENCE   LINES  175 

The  method  of  computing  stresses  for  any  train  of  loads  need  not  be  repeated  here 
except  to  call  attention  to  the  load  divides,  which  must  be  carefully  observed  for  each 
member  in  placing  the  loads  on  the  span. 

Temperature  effects.  For  a  general  case  of  unequal  temperatures  in  the  several  mem- 
bers the  following  assumptions  are  made: 

Let  the  normal  .  temperature  be  65°  F.,  for  which  the  structure  has  no  temperature 
stresses.  Then  assume  a  case  where  the  top  chord  is  heated  to  130°  F.;  the  bottom 
chord  to  104°  F.,  and  the  web  system  to  117°  F.  For  s=0.000007,  and  #=29000 
kips  per  sq.in.  the  values  of  Ed  become 

Top  chord  Z  =  130°-65°  =65°,  Eet  =  13.20; 
WebSystem  t  =  117°  -65°  -52°,  #rf  =  10.56; 
Bottom  chord  £  =  104°  -65°  -39°,  Eet=  7.92. 

The  values  E  Alt  =  dlE  and  ESaAlt  are  computed  in  Table  50A,  using  I  in  feet  and 
E  in  kip  feet.  Finally  the  half  sum  E2Sa4lt  is  found  to  be  3293.4  -3593.2  =  -299.8 
kip  feet.  Table  50fi  gives  ^E^lyw=54.7Q  and  from  Eq.  (50o)  for  cosa  =  l, 


-  *«.  *.       *  •—  •.     i     * 

«'==~=  ~5'  7°     PS' 


which  is  a  thrust  acting  outward  the  same  as  the  conventional  loading  Xa  = 
Hence  the  temperature  stress  in  any  member  becomes 


wherein  St  has  the  same  sign  as  Sa. 

A  more  severe  stress  would  be  produced  when  the  top  chord  is  colder  than  the 
bottom  chord,  or  when  the  top  chord  has  a  temperature  of  +10°  F.  at  the  same  time 
that  the  bottom  chord  has  a  temperature  of  -16°  F,  a  case  which  might  occur  on  a  clear, 
cold  day.  The  stresses  St  produced  by  such  a  condition  would  have  the  opposite  sign 
of  Sa  when  found  from  +Xat. 

For  a  uniform  rise  of  65°  above  the  normal,  the  elongation  for  the  whole  span 
becomes,  by  Eq.  (50E), 

_.!  A    D      771 

=  2226.6  ft., 


cos  a 
and  from  Eq.  (SOo), 


Edat      2226.6 


giving  stresses  St  =  —SaXai. 

For  a  uniform  fall  of  65°  below  the  normal  ^  ==  -2226.6  ft.  and  Xat  =  -20.33  kips, 

giving  stresses  St  =SaXat, 

Compare  these  results  with  those  of  the  example  in  Art.  59,  for  a  solid  web  arch  o 

about  the  same  dimensions. 


176  KINETIC  THEORY  OF   ENGINEERING  STRUCTURES  CHAP.  X 

Abutment  displacements.     Assuming  that  as  a  result  of  yielding  abutments  the  length 
of  span  is  increased  an  amount  J/=0.03  ft.  then  from  Eq.  (49L) 

0.03  X 29 ,000 _  _„         .  e 
•~  109.53  ''  °    lpS' 


Comparing  this  result  with  the  one  obtained  for  the  solid  web  arch  in  Art.  49,  it 
is  clear  that  the  framed  arch  is  almost  twice  as  stiff. 

Deflection  of  the  crown  due  to  the  temperature  effect  producing  Xat  =  —  5.48  kips, 
Table  50A,  and  Eq.  (50M).  In  this  case  the  abutments  are  assumed  rigid  making  Jr=0? 
and  the  stresses  S0  are  not  included,  since  temperature  effects  alone  are  desired.  This 
reduces  Eq.  (50c)  to 

i       /St  i  , 


where  St  =  -SaXat  and        =  -  =  -EAlXat;    also  Sl  =        for  a  load  unitY  at  the 

r  r  Zi 

center  of  a  symmetric  span,  and  dlE  =EAlt.     Hence 


which  can  easily  be  computed  for  any  Xat  using  the  values  SA)  EM  and  EAlt  given 
in  Table  50A. 

ART.  51.     TWO-HINGED   ARCH   WITH   CANTILEVER   SIDE    SPANS 

Occasionally  a  structure  of  this  type  is  peculiarly  adapted  to  certain  sites  as  the 
one  at  High  Bridge,  Ky.  Its  application  has,  however,  received  adverse  criticism  because 
the  analysis  of  stresses  was  considered  too  complicated. 

This  objection  might  apply  to  any  of  the  ordinary  methods  of  analysis,  but  not  when 
the  problem  is  solved  by  influence  areas. 

As  a  type  of  bridge  it  is  commendable  and  deserves  careful  consideration  whenever 
a  particular  site  offers  a  suitable  profile  and  good  foundations  for  the  center  span. 

Fig.  5lA  represents  one  of  several  forms  which  might  be  employed.  The  center 
span  is  the  same  as  the  arch  in  the  previous  article  with  the  exception  of  the  end  posts 
which  are  here  made  vertical,  thus  increasing  the  span  to  180  ft.  Owing  to  this  dif- 
ference, the  computations  in  Tables  50A  and  50s  no  longer  apply  because  the  stresses 
SA,  SB  and  Sa  are  now  slightly  changed.  Otherwise  the  preliminary  computations  would 
be  precisely  the  same  in  both  structures. 

The  difference  between  the  two-hinged  arch  and  the  present  structure  with  cantilever 
side  spans  is  entirely  in  the  principal  system  which  results  when  the  redundant  thrust 
Xa  is  removed.  In  the  first  case  the  principal  system  is  a  simple  truss  on  two  supports 
A  and  B,  while  in  Fig.  5lA  the  principal  system  becomes  a  cantilever  when  the  hinged 
support  at  A  is  made  movable  for  the  purpose  of  analysis.  The  supports  at  D  and  F 
are  roller  bearings  and  the  simple  trusses  DC  and  EF  are  hinged  at  C  and  E  respectively. 


ART.  51 


SPECIAL  APPLICATIONS  OF  INFLUENCE  LINES 


177 


178  KINETIC  THEORY  OF  ENGINEERING  STRUCTURES  CHAP.  X 

The  two-hinged  central  span,  as  in  any  ordinary  two-hinged  arch,  is  made  deter- 
minate by  substituting  a  roller  bearing  for  the  hinged  bearing  at  A. 

The  XQ  influence  line  is  found  as  in  the  previous  problem  and  the  computation  of  the 
Ew  loads  is  not  repeated  here.  However,  the  loads  Ew0  and_EVt-2,  must_now  be  included 
because  these  loads  affect  the  directions  of  the  extreme  rays  C'A'  and  B'E'  of  the  equilib- 
rium polygon  outside  of  the  span  AB  while  they  have  no  bearing  on 'the  A"a  influence 
line  within  this  span. 

The  complete  Xa  influence  line  for  the  whole  span  DF  is  found  by  prolonging  the 
extreme  rays  of  the  equilibrium  polygon  A'B'  to  the  points  C"  and  E'  and  finally  draw- 
ing the  lines  D'C'  and  E'F'.  The  middle  ordinate  A  may  be  computed  from  the  force 
polygon  as  follows: 


X=—    "       measured  to  scale  of  lengths, 

4/2 

or 

measured  to  scale  of  ordinates. 

The  lines  A'C'  and  B'E',  found  by  laying  off  A,  are  respectively  parallel  to  the 
extreme  rays  of  the  force  polygon. 

Stress  influence  areas-  These  are  found  precisely  as  in  Art.'  50,  so  far  as  the  span 
AB  is  concerned,  by  laying  off  end  ordinates  T,A  =&A/Sa  and  fjB=^B/Sa  and  the  factor 
becomes  a=Sa.  In  each  case  the  two  end  rays  so  determined  will  intersect  in  a  point 
ir,  which  is  vertically  under  the  center  of  moments  i  of  the  member  in  q^stion.  Also 
the  S0/Sa  influence  lines  must  be  straight  over  the  panel  containing  the  member. 

Outside  of  the  span  A'B  the  S0/Sa  lines-  are  drawn  as  for  any  cantilever  system  as 
per  Art.  26. 

ART.  52.     FIXED    FRAMED   ARCHES 

A  framed  arch  with  fixed  ends  has  three  external  redundant  conditions  according 
to  Eq.  (3c)  and  Fig.  3j,  and  hence  requires  for  its  analysis  three  elasticity  equations  either 
of  the  form  of  Eqs.  (7n)  or  Eqs.  (8D) . 

Temperature  changes  and  abutment  displacements  come  into  prominence  here. 
These  effects  are  bound  to  remain  more  or  less  problematic  because  the  actual  circum- 
stances attending  the  construction  and  later  life  of  the  structure  cannot  be  foretold 
with  any  high  degree  of  certainty.  Therefore,  it  would  seem  unnecessary  to  attempt 
the  analysis  of  the  load  effects  with  any  extraordinary  refinement  and  some  assumptions 
may  be  made  to  simplify  the  work,  provided  they  are  on  the  side  of  safety. 

It  is  nearly  always  permissible  to  neglect  the  effect  of  the  web  system  in  comput- 
ing the  elastic  loads  w.  In  preliminary  work  it  is  also  admissible  to  choose  a  constant 
cross-section  for  the  chord  members. 

As  a  general  criticism  it  might  be  added  that  fixed  framed  arches  are  not  commendable 
for  the  flat  type,  since  the  temperature  and  reaction  effects  produce  very  considerable 


ART.  52 


SPECIAL  APPLICATIONS  OF  INFLUENCE  LINES 


179 


stresses  which  increase  as  the  rise  diminishes.     Good  rock  foundations  must  be  available 
under  all  circumstances,  otherwise  the  fixed  arch  is  prohibitive. 

It  is  always  advisable  to  separate  the  computation  of  the  load  stresses  from  the 
temperature  and  reaction  displacement  stresses  so  as  to  determine  the  relative  importance 

of  the  latter. 

General  relations  between  the  external  forces  and  the  principal  system.  There  are  several 
wavs  in  which  the  three  external  redundant  conditions  may  be  applied,  depending  on 
the  choice  of  the  principal  system.  One  of  these  was  illustrated  in  Figs.  44E  to  J,  where 
the  principal  system  was  a  cantilever  fixed  at  one  end.  Another  assumption  might 
be  made  by  cutting  the  arch  at  the  crown  and  creating  two  cantilevers  fixed  at  the 
respective  abutments  of  the  arch. 

However,  the  simplest  determinate  structure  is  always  a  truss  on  two  supports, 
and  in  the  present  case  that  disposition  will  be  made,  as  it  affords  the  most  compre- 
hensive solution. 


FIG.  52A. 

All  three  methods  have  been  used  by  Professor  Mueller-Breslau  and  others,  and  the 
final  results,  are  identical  and  involve  about  the  same  amount  of  labor.     In  each  case 
the  external  redundant  conditions   may  be   so   chosen  that  the  application  of  the  « 
plified  Eqs.  (44A)  becomes  possible  in  accordance  with  the  discussion  in  Art   44 

Fi»    52A  illustrates  the  relation  of  the  external  forces  for  a  single  applied  load  I  , 
producing  reactions  R,  and  Rz,  intersecting  in  the  point  C  on  this  load.     This  fixes  the 
points  ttl  and  b,  on  the  verticals  through  the  outer  supports  a  and  ftjnth  span  I.     ine 
triangle  ^Cb[  thus  becomes  a  resultant  polygon  with  the  closing  line  a^. 
Rl  may  be  resolved  into  the  vertical  component  A0  and  the  haunch  thrust  H   along  afr. 
The  reaction  R2  may  similarly  be  resolved  into  the  vertical  reaction  B0  and  the 
H'   which  latter  is  equal  and  opposite  to  the  H'  acting  at  ai. 

'   The  vertical  reactions  A0  and  B0  are  the  same  as  for  a  simple  beam  o*  with  deter- 
minate supports  A0  and  B0.     Hence  A0=P(l-e)+l  and  B0  =  Pe/l.     Also  the  moment,  f, 
P  intTof  the  simple  beam,  equals  M^-KH,  wh_ere_H  is  the  honzonta  componen 
'  and  K  is  the  vertical  ordinate  of  the  triangle  a.Cb,  through  the  point  m.     Hence 


180 


KINETIC  THEORY  OF   ENGINEERING   STRUCTURES 


CHAP.  X 


K  =  Mom/H=Mom/H'  cos  a.  Therefore,  the  resultant  polygon  alCbl  is  determined 
when  H'  and  the  closing  line  a^bi  are  found. 

The  external  redundant  conditions  are  now  removed  by  relieving  the  fixed  ends 
and  resting  the  curved  structure  on  two  determinate  supports  at  the  extreme  outer 
points  a  and  b,  Fig.  52B.  This  simple  truss  of  span  /,  provided  with  a  hinged  bearing  at 
a  and  a  roller  bearing  at  b,  constitutes  the  principal  system. 

To  each  end  of  the  principal  system  an  imaginary  rigid  disk  is  attached  as  shown 
in  the  figure  by  the  two  shaded  triangles.  The  redundant  conditions  are  now  applied 
to  these  disks  as  external  forces  or  moments,  thereby  re-establishing  conditions  of  stress 
in  the  principal  system  which  are  identical  with  those  produced  in  the  original  structure 
while  the  redundant  conditions  were  active.  The  two  disks  are  not  connected  at  0,  but 
are  free  to  transmit  a  set  of  forces  independently  to  each  abutment. 

The  equal  and  opposite  forces  Xb  are  chosen  vertically,  and  the  one  acting  upward 
is  supposed  to  act  on  the  disk  OA.  The  forces  Xc  are  equal  and  opposite,  but  of  unknown 
direction  /?  with  the  horizontal,  the  one  acting  to  the  right  is  applied  to  the  disk  OA.  The 


FIG.  52s. 

two  moments  Xa  are  also  equal  and  opposite  and  act  separately,  one  on  each  disk.  The 
pole  0  is  not  yet  determined,  but  will  be  fixed  by  certain  geometric  conditions  to  be 
established  later. 

To  determine  the  exact  relations  between  these  redundant  forces  and  the  principal 
system  it  is  preferable  to  discuss  only  the  forces  acting  on  one  end  of  the  span.  In  Fig. 
52c,  the  left-hand  abutment  is  shown  with  the  redundant  forces  which  are  active  at 
that  end  only.  A  similar  set,  not  shown,  would  be  active  at  the  right-hand  abutment. 

The  structure  is  now  referred  to  coordinate  axes  (x,  y)  with  origin  at  0.  The  y  axis 
is  made  vertical,  and  the  x  axis  is  coincident  with  the  redundant  Xc  making  the  angle 
/?  with  the  horizontal.  The  location  of  0  and  the  angle  /?  are  still  unknown.  The  ordi- 
nate  ym  of  any  point  m  is  measured  vertically  from  the  x  axis,  while  the  abscissa  xm  of 
this  point  is  measured  horizontally  from  the  y  axis  instead  of  parallel  to  the  x  axis.  This 
is  more  convenient Jnjihe  considerations  which  follow. 

The  rigid  disk  aa'O  connects  the  origin  0  with  the  arch  along  aa' ',  and  to  this  origin 
are  applied  two  equal  and  opposite  forces  H' ,  which  are  equal  and  parallel  to  the  original 
haunch  thrust  acting  at  a^  The  equilibrium  of  the  principal  system  and  of  the  fixed 


ART.  52 


SPECIAL  APPLICATIONS  OF  INFLUENCE  LINES 


181 


arch  thus  remains  undisturbed.  In  the  original  fixed  condition  the  force  P  produced 
the  reactions  RI  and  R2  and  these  were  resolved  into  components  A0  and  H'  acting  at 
ai  and  B0  and  H'  acting  at  61. 

Suppose  now  that  all  the  external  forces  to  the  left  of  a  section  tt  act  on  the  principal 
system  only,  and  that  the  three  forces  H'  and  the  vertical  reaction  A0  are  applied  to  the 
rigid  disk  aa'O  and  are  thence  transmitted  to  the  principal  system.  Then  the  force  H' 
at  a\  and  the  force  H' ',  in  opposite  direction  at  0,  form  a  couple  with  lever  arm  z0  cos  a 
producing  a  moment  Xa=H'z0cos  a.  The  other  force  H',  acting  at  0  and  to  the  right, 
may  be  resolved  into  two  components  Xb  and  Xc,  where  Xb  is  vertical  along  the  y  axis, 
and  Xc  acts  along  the  x  axis.  The  external  forces  to  the  left  of  the  section  and  acting 
on  the  principal  system,  are  then  P,  A0,  Xb,  Xc  and  a  moment  Xa=H'z0  cos  a  =Hz0. 
Of  these  the  two  forces  Xb  and  Xc  and  the  moment  Xa  constitute  the  redundant  condi- 
tions while  the  forces  P  and  A0  are  known  and  all  are  applied  to  the  principal  system 


FIG.  52c. 


to  the  left  of  the  section  tt.    A  similar  set  of  external  forces  acts  on  the  principal  system  to 
the  right  of  the  section,  but  these  are  not  shown  in  Fig.  52c. 

The  moment  of  all  external  forces  about  any  point  m  of  any  frame,  involving  thre 
redundant  conditions,  is  expressed  by  Eq.  (7  A)  as 


Mm=Mom-MaXa-MbXb-McXc 


(52A) 


wherein  Mom=A0(li-xm) -Pd-^ the  moment  about   m  due  to  the  load  P  acting  on  the 
principal  system.     This  is  condition  X=0.  .     . 

Ma  =  l=the  moment  about  m  due  to  the   moment  Xa  =  l  applied  to  the  principal 

system.     Condition  Xa  =1.  .     .     , 

M6  =  l.*m=the  moment  about  m  due  to  the  force  Xb  =  l  acting  on  the  prmcipa 

system.     Condition  Xb  =  l. 

Mc  =  l.</mcos/?=the  moment  about  m  due  to  the  force  Xc  =  l  acting  on  the  prin- 
cipal system.     Condition  Xc  =  l. 


182  KINETIC  THEORY  OF   ENGINEERING  STRUCTURES  CHAP,  x 

Substituting  these  values  into  Eq.  (52  A)  gives  the  following  fundamental  moment 
equation  for  fixed  arches: 

Mm=Mom  —  l-Xa—xmXb—ym  cos8Xc  ......     (52s) 

Before  Mm  can  be  determined  for  any  point  m  of  the  arch,  the  three  redundants 
Xa,  Xb  and  Xc  must  be  evaluated  from  three  simultaneous  work  equations  of  the  form 
of  Eqs.  (44n)  ,  which  equations  may  be  made  to  apply  to  the  present'  problem  by  so 
locating  the  (x,  y}  axes  that  dab,  dac  and  dbc  all  become  zero. 

Since  there  are  as  many  values  for  the  redundants  as  there  are  panel  points  on  the 
arch,  and  for  each  point  there  will  be  as  many  values  as  there  are  positions  of  the  mov- 
ing loads  requiring  investigation,  the  only  practical  solution  of  the  problem  is  by  means 
of  influence  lines,  first  for  the  three  redundants  and  then  for  the  moment  Mm  for  each 
member  of  the  arch. 

The  stress  S  in  any  member  may  always  be  found  from  the  moment  Mm  about 
the  center  of  moments  m  for  that  member  and  the  lever  arm  rm  measured  from  this  moment 
center  perpendicular  to  the  direction  of  the  member.  Hence 

„    Mm  Mom  Ma  Mb  Mt 

o  =  -  ;       o0=  ---  ,       ^a—  --  >       &b  —  —  ,       ana       oc  =  —  ,    .     .     .     (o2c) 

I'm  I'm  ?"m  ^m  I'm 

where  the  lever  arm  rm  is  constant  for  the  same  member. 

Therefore,  Eq.  (52s)  will  furnish  the  stress  in  any  member  as 

M        1 
S  =—  -  =—  [Mom  -l-Xa  -xmXb  -ym  cos  pxe]  =S0  -SaXa  -SbXb  -SCXC,         (o2c) 

'  m        'm 

„      Ma      I         „      Mb     xm  „      Mc     ym  cos  8 

where       Sa  =  —  -  =  —  ;      Sb=—  ?=  —  ;       and       Sc  =  —  =if—  -  E  .......     (52E) 

*•  *•       -  ««'  4*  4"  V  / 

'  m       'm  'm        'm  'm  'm 

Location  of  the  coordinate  axes.  According  to  the  conditions  imposed  by  the  simplify- 
ing process  of  Eqs.  (8n),  discussed  in  Art.  44,  the  coordinate  axes  must  be  so  located 
that  the  displacements  d,  bearing  different  double  subscripts,  must  be  made  zero. 

These  displacements  d  as  given  by  Eqs.  (SB),  then  become 

=0', 


and  substituting  the  values  for  the  stresses  from  Eqs.  (52E)  then 

^    xl        v  1 

—  ^  2j  XW    =  (J 


.       ^xycosfll  v 

=04,  =  1  -^f          =  cos  P2xywa  =0 


(52r) 


ART.  52  SPECIAL  APPLICATIONS  OF   INFLUENCE  LINES  183 

where  wa  =  l-l/EFr2  for  a  moment  unity,  according  to  Eqs.  (36B)  and  (36o),  giving 
wa=4l/r=Sl/EFr=Ml/EFr2  and  representing  a  certain  geometric  function  called  an 
elastic  load  for  some  particular  pin  point  with  coordinates  (x,  y). 

Eqs.  (52r)  then  represent  the  conditions  which  determine  the  location  of  the 
coordinate  axes  such  that  the  simple  work  Eqs.  (44A)  become  applicable. 

The  first  two  conditions  (52r)  imply  that  the  origin  0  of  the  (ar,  ?/)  coordinate  axes 
is  the  center  of  gravity  of  the  several  elastic  loads  of  all  panel  points  of  the  principal  sys- 
tem. This  must  be  so  because  the  moments  ^lxwa  and  %ywa  could  not  be  zero  unless  the 
resultant  Hwa  passes  through  the  origin.  According  to  the  third  condition,  the  angle 
between  the  axes  must  be  such  that  the  centrifugal  moment  2>xywa=Q,  which  is  true 
when  the  axes  are  related  as  conjugate  axes. 

The  origin  0  may  then  be  located  with  respect  to  any  assumed  pair  of  axes  as  the 
(z,  v)  axes  in  Fig.  52c,  where  the  z  axis  is  taken  vertically  through  the  point  a  and  the  v 
axis  is  any  horizontal  axis  conveniently  located  say  through  a'.  The  coordinates  z,  i\ 
of  all  the  panel  points,  are  then  determined  from  the  arch  diagram  and  tabulated.  The 
wa  elastic  loads  are  computed  from  Eq.  (36fl)  (and  Eqs.  (36D)  if  the  web  members  are 
included)  for  each  pin  point.  See  the  problem  in  Art.  51.  Finally  the  moments  zwa 
and  vwa  are  found  and  from  these  the  coordinates  li  and  z0'  for  the  origin  0  are  obtained 

from 

,       ^vwa  ,     2zwa  ._0  , 

''-        and  *  ~     ..... 


This  fixes  the  y  axis,  which  is  parallel  to  the  vertical  z  axis  through  the  center  of 
gravity  0.  The  x  axis,  while  passing  through  0,  makes  some  angle  ft  with  the  horizontal 
such  that  I,xywa=0,  according  to  the  last  of  Eqs.  (52r). 

The  (x,  y)  coordinates  are  derived  from  the  (z,  v)  coordinates  when  the  angle  /?  is 
determined.  Taking  /?  positive  when  measured  to  the  left  of  the  origin  and  below  the 
horizontal,  or  to  the  right  and  above  the  horizontal  as  shown  in  Fig.  52c,  then 

x=li—  v  and  y=z  —  z0'  +x  tan  /?  ......     (52H) 

The  angle  0  is  found  by  substituting  the  value  for  y  from  Eqs.  (52n)  into  the 
condition  equation,  giving 

—z0f  +x  tan  fi]wa  =0; 


or  I>xzwa  -z0fZxwa  +  tan  p2x2wa  =0; 

and  noting  that  2xwa=Q,  then 

(52j) 


The  abscissa?  x  being  known  from  Eqs.  (52n)  the  values  Xxzwa  and  Xx2wa  are  readily 
found  and  tan  /?  is  then  obtained  from  Eq.  (52j).  Finally,  the  y  ordmates  are  computed 
by  the  second  Eq.  (52n)  and  the  new  axes  and  (x,  y)  coordinates  are  thus  determined. 

For  symmetric  arches  the  x  axis  is  horizontal  and  tan  £=0,  thus  greatly  lessening 
the  foregoing  computations. 


184 


KINETIC   THEORY  OF  ENGINEERING  STRUCTURES 


CHAP.  X 


Influence  lines  for  Xa,  X&,  Xc  and  Mm.  The  coordinate  axes  (x,  y)  having  been  located 
to  fulfill  the  requirements  making  ^=^=0,  dac=dca=0  and  dbc=ocb=0,  the  simplified 
Eqs.  (44A)  become  applicable  to  the  present  problem. 

Since  it  is  desirable  to  investigate  separately  the  effects  due  to  loads,  temperature 
and  abutment  displacements,  the  load  effect  will  be  considered  first  for  immovable  abut- 
ments.    Eqs.  (44A)  then  take  the  simple  form  of  Eqs.  (44B),  as 
YP  ft 

*^*-    m.u tnn.  -*T  — -«.    jji^  jnu  i  -\r  *-*  *•    m^  jnr.  /  ~c\      \ 

(,)2K) 


and 


Y  p  ^ 

*->*  mumc 

r s  • 


'» 


Eqs.  (SB)  give  the  displacements  dm,  £w>  and  £cc  in  terms  of  the  stresses  in  the 
members.  Noting  that  p  =l/EF,  and  substituting  the  values  for  Sa,  Sb  and  Sc  from  Eqs. 
(52E) ,  then  Eqs.  (SB)  give 


EFr2' 


» 
dbb  = 


'EFr2' 


and  calling  l/EFr2=wa;   x2l/EFr2=x2wa=xwb  and  y2l/EFr2  =ij2wa=ywc  then  Eqs.  (52x) 
become  for  a  single  load  unity  at  any  point  ra : 


1  •  $mb        1  ' ' 


*•  '  " 


'  " 


(52L) 


Eqs.  (52L)  furnish  the  values  Xa,  Xb  and  Xc  for  any  position  m  of  a  single  moving 
load  PTO  =  1,  as  functions  of  the  deflections  dma,  d^  and  dmc  of  the  point  m  resulting  from 
the  conventional  loadings  Xa  =  ],  Xb  =  l,  and  Xc  =  l. 

The  exact  significance  of  the  deflections  3ina,  dmb  and  dmc  may  be  determined  from 
their  values  as  given  by  Eqs.  (8 A),  noting  that  S0  =  Mom/rm  for  a  load  P^-l  applied  at 
m  from  Eqs.  (52c),  also  using  the  values  given  by  Eqs.  (52E).  Hence 

is>  v^oo      '     y  "•*•  om' 


EF 

J_ 

blEF 

I 


Momxl 


=cos 


(52M) 


Now  ^ma  is  the  deflection  of  a  point  m  due  to  Xa  =  l  and  PTO  =0,  and  by  Eqs.  (52M) 
it  is  equal  to  the  sum  of  the  moments  Momwa  of  the  loads  wa  about  m.  Hence  dma  must 
represent  the  ordinate  to  a  moment  digram  drawn  for  the  loads  wa  with  a  pole  unity. 
Also,  if  this  moment  diagram  be  drawn  with  a  pole  Ha  =  2wa,  then  the  resulting  ordinates 
i)a  will  represent  ordinates  of  the  Xa  influence  line  according  to  Eqs.  (52L). 


ART.  52  SPECIAL  APPLICATIONS  OF  INFLUENCE  LINES  185 


Similarly  the  moment  diagram  drawn  for  the  loads  wb=xwa  with  a  pole  Hb  = 
will  furnish  ordinates  jjb  for  the  Xb  influence  line  and  a  third  moment  diagram  drawn 
'or  the  loads  wc  =ywa  with  a  pole  Hc  =  cos  {32>ywc  will  give  ordinates  )jc  for  the  Xc  influence 
ine,  noting  that  one  of  the  cos  /?  factors  cancels. 

Hence  the  equations  for  the  influence  lines  of  the  three  redundant  conditions  from 
Eqs.  (52L),  become 

1      /^  <> 


(52N) 

dr, 


me 


TTvi  TT 

cos    Lwc    H 


*•-< 

These  influence  lines  remain  the  same  for  the  same  structure  and  hence  are  drawn 

only  once. 

The  moment  influence  line  Mm,  for  any  point  TO  with  coordinates  xm  and  ym,  is  derived 
from  Eq.  (52s) ,  giving  any  moment  ordinate  as 

r,m=Mm=l\lf>m-\Xa+xmXb+ymcos8Xc]=rjom-[-na+xm-f)b+ymcos  fa],     .     (52o) 


where  Tlom  is  any  ordinate  of  the  ordinary  moment  influence  line  Mom,  drawn  for  the  point 
TO  of  a  simple  beam  on  two  determinate  supports  A0  and  B0,  which  is  a  different  line  for 
each  point  TO.  The  ordinates  r)a,  yb  and  TJC  are  those  respectively  of  the  Xa,  Xb  and  Xc 
influence  lines  all  under  the  same  panel  point  of  the  truss. 

Hence,  the  moment  influence  line  Mm  for  any  point  TO  is  drawn  by  computing  the 
ordinates  -[rja+xmrjb+ym  cos  fa]  for  _all  panel  points  of  the  loaded  cord  and  plotting 
these  ordinates  negatively  from  the  Mom  influence  line.  Positive  areas  thus  correspond 
to  positive  moments. 

This  Mm  influence  line  will  serve  to  find  the  maximum  and  minimum  bending 
moments  for  the  point  TO  due  to  any  system  of  concentrated  loads.  The  influence  line 
gives  the  load  divides. 

The  stress  influence  line  for  any  member  may  be  derived  from  the  moment  influence 
line  drawn  for  the  center  of  moments  TO  for  that  particular  member.  Thus  if  rm  is  the 
lever  arm  for  a  certain  member  S  with  center  of  moments  m,  then  S=Mm/rm,  and  the 
ordinates  T?,  for  this  stress  influence  line  by  Eq.  (52o)  become 


(52P) 


where  the  several  9  ordinates  are  as  above  defined  and  all  measured  under  the  same 
panel  point.  Positive  areas  then  give  positive  or  tensile  stress.  The  moment  influence 
line  for  the  center  of  moments  for  any  member  may  be  directly  used  by  applying  the 
factor  l/rw  to  obtain  the  stress. 

The  rather  lengthy  operation  of  computing  all  these  ordinates  for  the  several  stress 
influence  lines  for  all  the  members  may  be  considerably  shortened  by  employing   the 


1S6 


KINETIC  THEORY  OF   ENGINEERING  STRUCTURES 


CHAP.  X 


following  suggestions:  Thus  the  multiplications  xm-fjb  and  ym  cos  /?j?c,  may  be  performed 
graphically  by  laying  off  angles  whose  tangents  are  respectively  xm  and  ym  cos  /?,  as  illus- 
trated in  Fig.  52o,  using  the  scale  of  the  TJ  ordinates. 
Two  such  diagrams  are  required  for  each  Mm  or  S  line. 

If  a  proportional  divider  is  at  hand,  then  it  may  be 
used  to  perform  these  multiplications  by  setting  the  divi- 
ders first  for  the  ratio  1 :  xm  and  then  for  1 :  ym  cos  /?. 

Aside  from  the  above  methods,  the  stress  influence 
line  for  any  web  member  may  be  derived  from  the 
influence  lines  of  two  adjacent  chords  by  using  the 
following  method  given  by  Professor  Mueller-Breslau. 

When  the  top  chord    is   the    loaded  chord,  then  the 
FIG.  52o.  adjacent  members  of   the   bottom   chord  are  used,  and 

vice  versa. 

In  Fig.  52E,  assume  a  unit  positive  stress  in  L\  and  resolve  the  same  into  com- 
ponents— £1  parallel  to  D\  and  +£2  parallel  to  D2. 


FIG.  52E. 


FIG.  52r. 


Then  assume  a  unit  positive  stress  in  L2  and  resolve  this  into  components  +  n\ 
parallel  to  D\  and  — /i2  parallel  to  D2.  Then  for  the  top  chord  loaded  and  no  load  at 
tr>,  the  stresses  D\  and  D2  become 


D2=+e2Ll  - 


+L2 


(52Q) 


By  substituting  influence  ordinates  for  the  stresses  D  and  L  this  furnishes  a  ready 
means  of  deducing  a  D  influence  line  from  the  two  influence  lines  of  the  adjacent  chords. 
Thus  the  DI  influence  area  is  the  area  between  the  L2  line  and  the  z\L\/ //t  line  and  the 


ART.  52 


SPECIAL  APPLICATIONS  OF  INFLUENCE  LINES 


187 


final  Di  area  thus  obtained  will  have  a  factor  m.     The  ordinates  to  be  used  in  the  equa- 
tions are  always  under  the  same  load  point. 

When  the  bottom  chord  is  the  loaded  chord  and   there  is  no  load  at  the   point   n, 
then  using  the  stresses  as  indicated  in  Fig.  52F,  the  diagonals  D2  and  D3  become 


(52  R) 


The  value  of  D2  will  be  the  same  in  both  equations  Q  and  R  and  the  signs  will  come 
alike  by  noting  that  the  influence  ordinates  for  the  L  and  U  lines  have  opposite  signs  and 
hence  the  coefficients  s  and  «  have  opposite  signs  from  those  used  in  Eqs.  (52Q). 

The  resultant  polygon  for  any  case  of  applied  loads  P  may  be  located  by  finding  the 
corresponding  values  of  Xa,  Xb,  and  Xc  from  the  three  influence  lines  for  thes 
having  previously  located  the  (a;,  y)  axes. 

The  redundants  Xe  and  Xb,  Fig.  52c?  were  taken  as  the  components  of  H  ,  respec- 
tively coincident  with  the  x  and  y  axes.     The  angle  /?,  which  the  x  axis  makes  with  t 
horizontal  is  given  by  Eq.  (52j)  and  the  y  axis  was  taken  vertically, 
izontal  component  H  of  Xc  is 

H  =XC  cos  /?,       and  from  Fig.  52c, 
tan  a.=t&n  (3+-jj 


cos  a 


z°~ 


cos  a 

Xa  =  Xg 

H'cosa      H 


H 


.     .     (52s) 


Xa  = 


COS      = 


These  dimensions  fix  the  location  of  the  closing  line  a 
of  the  abutments  also  the  haunch  thrust  H'   all  in  terms  of 


on  the  two  end  verticals 
X   and  X 


r, 


188  KINETIC  THEORY   OF  ENGINEERING  STRUCTURES  CHAP.  X 

A  force  polygon  is  drawn  by  laying  off  all  the  loads  P  in  proper  succession,  dividing 
this  load  line  into  the  parts  A0  and  B0  and  at  the  dividing  point  draw  a  line  parallel  to 
a[bi  of  length  equal  to  H' '.  This  determines  the  pole  of  the  force  polygon  from  whicli 
the  reactions  RI  and  R2  and  the  resultant  polygon  through  ai  and  61  are  easily  drawn. 
See  also  the  force  polygon  in  Fig.  52A,  showing  how  the  pole  0  is  located  when  H' ,  A0 
and  B0  are  given. 

Temperature  stresses.  For  the  general  case  of  temperature  effects,  each  member  may 
be  supposed  to  undergo  some  change  ±4lt=etl  in  its  length.  These  changes  will  pro- 
duce a  deformation  in  the  frame,  giving  rise  to  external  redundant  forces  Xat,  Xbt  and  Xct, 
which  may  be  found  from  Eqs.  (44c),  where  abutment  displacements  and  load  effects 
are  excluded. 

Using  the  values  of  dat,  $bt  and  dct  as  given  by  Eqs.  (Sc)  and  then  noting  the  sub- 
stitutions for  Saa  =  £wa,  dbb  =  2>xwb  and  dcc  =  *£>ywc  made  in  Eqs.  (52N) ,  the  Eqs.  (44c)  become 


Y     _dat_'£SaJlt 

-A  at * vi 

Oaa 

_dbt_ 

J\.bt — ^ 


COS 


(52T) 


where  dat  represents  a  rotation  measured  in  arc,  while  dbt  and  dct  are  the  displacements 
of  the  origin  0,  due  to  the  assumed  temperature  changes  measured  in  the  directions  of  the 
y  and  x  axes  respectively. 

These  redundant  conditions  would  then  produce  a  moment  Mmt  about  any  point 
m  as  given  from  Eq.  (52s),  where  Mom=Q,  thus 

-Mmt=Xat+xmXbt+ymcos^Xct.        .....     (52r) 

Since  the  changes  Alt  in  the  lengths  of  the  members  represent  a  simultaneous  condition 
and  the  stresses  Sa,  Sb  and  Sc  were  previously  found  in  computing  the  w  loads,  it  is  best 
to  solve  Eqs.  (52T)  analytically,  using  the  denominators  previously  found  for  the  pole 
distances  of  the  X  influence  lines.  The  moment  Eq.  (52u)  can  then  be  solved  for  any 
moment  point  to  obtain  the  stress  in  .a  corresponding  member. 

For  a  uniform  change  in  temperature  of  ±  t°  from  the  normal,  an  approximate  deter- 
mination of  temperature  stresses  may  be  obtained  for  arches  of  high  rise.  For  flat  arches 
this  assumption  would  not  be  permissible.  When  the  effect  of  the  redundants  Xa  and 
Xb  is  neglected  then  the  stresses  may  be  found  from  Eqs.  (52T)  and  (52u)  as 


v  --    ........     (52v) 

\ 


The  effect  of  abutment  displacements  may  be  investigated  in  a  similar  manner,  using 
Eqs.  (44A)  and  omitting  the  terms  representing  the  loads  P  and  the  temperature.     The 


\BT.  52  SPECIAL  APPLICATIONS  OF  INFLUENCE  LINES  189 

iisplacements  Ar  must  then  be  assumed  or  estimated  and  oa,  db  and  dc  become  zero.  Hence 
the  redundants  and  moments  may  be  found  in  precisely  the  same  manner  above  illustrated 
for  temperature  effects,  thus 


Xar  — 


**. 

(52w) 


where  the  d's  in  the  denominators  are  those  used  in  Eqs.  (52T). 

Example.  Owing  to  the  comparatively  few  fixed  arches  of  the  framed  type  in 
existence,  it  was  difficult  to  find  a  suitable  example  to  use  in  illustrating  the  above  method 
of  analysis.  For  this  reason  the  present  example,  Figs.  52a,  was  taken  with  slight  modi- 
fication, from  the  one  given  by  Professor  Mehrtens  in  his  "  Statik  der  Baukonstruk- 
tionen,"  Vol.  Ill,  p.  343. 

The  bridge  has  a  clear  span  of  162  ft.  and  rise  of  23.69  ft.  The  roadway  is  30  ft. 
wide  and  the  estimated  weight  is  about  400,000  Ibs.,  making  20  kips  per  truss  per  panel. 
The  uniform  live  load  is  taken  at  110  Ibs.  per  sq.ft.  of  roadway  or  30  kips  per  truss  per 
panel,  for  medium  highway  loading.  The  top  chord  is  the  loaded  chord  and  the  arch 
is  symmetric,  making  /?— 0. 

"  The  top  chord  panel  points  are  on  the  arc  of  a  circle  whose  radius  is  Ru  =  163.7  ft. 
The  chords  are  parallel  and  the  radius  for  the  bottom  chord  points  is  Rt  =  148.7  ft. 

Ordinarily  the  computation  might  be  carried  out  by  neglecting  the  effect  of  the 
web  members,  but  for  the  sake  of  completeness  all  members  will  be  included. 

The  cross-sections  F,  lengths  I,  and  lever  arms  r  of  all  the  members,  are  tabulated 
in  Tables  52A  and  52n,  and  the  Ewa  loads  are  computed  by  successive  steps  indicated  by 
the  headings  of  the  columns.  The  lever  arms  ru  and  rn  are  found  as  described  in  Figs. 
36B  and  36F  In  the  present  solution  the  areas  were  converted  into  sq.ft.  in  computing 
Al  so  that  the  modulus  E  enters  with  its  real  value  of  4,176,000  kips  per  sq.ft.  instead  of 
29  000  kips  per  sq.in.  as  used  in  the  problem  of  Art.  51.  The  stresses  Sa  due  to  a  moment 
fc-1  kip  ft.,  are  easily  found,  as  the  reciprocals  of  the  lever  arms  r,  being  careful  1 
observe  the  signs  of  the  stresses. 

The  arch  being  symmetric,  the  y  axis  is  known  to  be  the  vertical  axis  of  symmetry, 
and  the  angle  ft,  which  the  x  axis  makes  with  the  horizontal,  becomes  zero.  Hence,  the 
(x  y)  axes  are  located  by  computing  the  ordinate  z0' =E2zwa/EXwa-.21.927  *•;  thus 
determining  the  distance  from  the  v  axis  to  the  center  of  gravity  0  of  the  wa  loads,  bee 
Table  52c.  The  y  ordinates  then  become  y=z-z0'.  The  (x,  z)  coordinates  must  of 
course  be  determined  from  the  arch  dimensions  or  the  equations  of  the  chord  curves. 
Table  52c  then  gives  the  functions  Ewb=Exwa,  Ewc=Eywa,  Exwb  and  Eywc  all  in 
terms  of  the  wa  loads.  The  sums  EXwa,  EXxwb  and  E2ywe  represent  the  pole  distances 
for  the  Xa,  Xb  and  Xe  influence  lines  such  that  the  influence  ordinates  will 
when  measured  to  the  scale  of  lengths. 


190 


KINETIC  THEORY  OF  ENGINEERING   STRUCTURES 


CHAP.  X 


Xc  INFLUENCE  LIKI^-ORDINATES  IOO  TIMtSfO  SCALE  OF  LENGTHS 


NCE  LINE-ORDIMATCSACTUALTOSCALEOFLfNGTHS.    ! 
I 


FIG.  52c. 


SPECIAL  APPLICATIONS  OF  INFLUENCE  LINES 


191 


TABLE  52A 
wa  LOADS   FOR  CHORD  MEMBERS 


l 

.S',, 

Sal 

Chord. 

F 

I 

r 

*EJl  =  jr 

*Ewa  =  ^ 

Panel 
Point. 

Sq.in. 

ft. 

ft. 

Kips. 

Kips  sq.ft. 

ft. 

0   2 

100 

20.73 

14.92 

-0.0670 

-0.0965 

-2.002 

0.1341 

1 

2  4 

110 

19.51 

14.84 

-0.0674 

-0.0883 

-1.728 

0.1165 

3 

4  6 

120 

18.72 

14.79 

-0.0676 

-0.0811 

-1.512 

0  .  1022 

5 

fi  8 

124 

18.25 

14.77 

-0.0677 

-0.0786 

-1.440 

0.0975 

7 

8  10 

130 

18.03 

14.79 

-0.0676 

-0.0737 

-1.325 

0.0896 

9 

82 

20.59 

15.14 

0.0661 

0.1161 

2.390 

0.1580 

2 

Q     K 

94 

19.33 

15.21 

0.0657 

0.1008 

1.944 

0.1277 

4 

^    7 

100 

18.55 

15.25 

0.0656 

0.0945 

1.757 

0.1152 

6      . 

7  9 

104 

18  .  13 

15.28 

0.0654 

0.0906 

1.642 

0.1074 

8 

1(9-9') 

128 

9.00 

15.31 

0.0653 

0.0734 

0.662 

0.0432 

10 

*  Above  wa  loads  are  for  £'  =  29,000X144=4,176,000  kipa  per  sq.ft. 

TABLE  52s 
wa  LOADS   FOR  WEB   MEMBERS 


Member. 

F 
Sq.in. 

I 
ft. 

r 

ft. 

*-£ 

Kips. 

Sa 
f-p 

£ips.  sq.ft. 

.<s  ; 
tUl-^s 

r 

ft. 

ru 
and 
rn 

ft. 

*Edl 

*  Ewa  =  

For  loaded  panel 
Points. 

At 
Panel 
Point. 

+• 

- 

0-1 
1-2 

2-3 
3^ 

4-5 
5-6 
6-7 
7-8 
8-9 
9-10 

8 
14 

18 
10 
18 
6 
14 
6 
10 
6 

15.03 
24.06 

15.26 
21.79 
15.66 
20.09 
16.19 
18.78 
16.91 
17.76 

780.87 
132.91 
443.54 
154.04 
331.25 
175.75 
260.26 
206.93 
231.40 

+  0.00128 
0.00752 
0.00225 
0.00649 
0.00302 
0.00569 
0.00384 
0.00483 
0.00432 

+  0.0132 
0.0602 
0.0324 
0.0519 
0.0725 
0.0585 
0.0922 
0.0696 
0.1037 

+  0.3176 
0.9186 
0.7060 
0.8128 
1.4565 
0.9471 
1.7315 
1  .  1769 
1.8417 

{12.81 
\13.05 
/21.90 
\18.96 
[13.38 
\13.65 
J19.98 
\17.70 
/  13.80 
1  14.40 
/18.45 
\16.62 
/14.37 
\15.18 
/17.10 
\15.84 
J15.03 
\16.05 

0.0248 

0 
2 
2 
4 
2 
4 
4 
6 
4 
6 
6 
8 
6 
8 
8 
10 
8 
10 

0.0243 
0.0419 

0.0485 
0.0533 

0.0517 
0.0407 

0.0459 
0.1055 

0.1011 
0.0513 

0  .  0570 
0.1205 

0.1141 
0.0688 

0.0743 
0  .  1225 

0.1148 

* 

Above  w.  loads  are  for  £  =  29,000X144  =  4,176,000  kips  per  sq.it. 

192 


KINETIC  THEORY  OF  ENGINEERING  STRUCTURES 


CHAP.  X 


TABLE  52c 
COMPUTATION  OF  w  LOADS  AND   POLE  DISTANCES 


T/it  <»1 

Pin 

Points. 

own 
Ewa 
Loads. 

z 

Ezwa 

X 

y=z-z0' 

Exwa 

Ewc  = 

Eywa 

Exwb 

Eywc 

Pin 
Points. 

ft. 

ft. 

ft. 

0 

0.0248 

12.03 

0.2983 

90 

-   9.897 

2.232 

-0.245 

200.88 

2.429 

0 

1 

0.1341 

0.00 

0.0 

81 

-21.927 

10.862 

-2.940 

879  .  83 

64.474 

1 

2 

0.1451 

22.31 

3.2372 

72 

+   0.383 

10.447 

+  0.056 

752  .  20 

0.021 

2 

3 

0.1165 

9.99 

1.1638 

63 

-11.937 

7.340 

-1.391 

462.39 

16.601 

3 

4 

0.1893 

29.83 

5.6468 

54 

+   7.903 

10.222 

+  1.496 

552  .  00 

11.823 

4 

5 

0.1022 

17.03 

1.7405 

45 

-  4.897 

4.599 

-0.501 

211.55 

2.451 

5 

6 

0.1292 

34.99 

4  .  5207 

36 

+  13.063 

4.651 

+  1.688 

167.44 

22  .  046 

6 

7 

0.0975 

21.53 

2.0992 

27 

-   0.397 

2.633 

-0.039 

71.08 

0.015 

7 

8 

0.1040 

38.01 

3.9530 

18 

+  16.083 

1.872 

+  1.673 

33.70 

26.900 

8 

9 

0.0896 

23.69 

2  .  1226 

9 

+    1.763 

0.806 

+  0.158 

7.26 

0  .  279 

9 

4(10) 

0.0027 

39.00 

0.1053 

0 

+  17.073 

0.0 

+  0.046 

0.00 

0.787 

4(10) 

B 

-2w?a  = 

E 

-2zwa  = 

E 

E 

E 

E 

2 

2 

2 

2 

1  .  1350 

24  .  8874 

55.664 

+  5.117 

3338.33 

147.826 

24  .  8874 

-5.116 

H  o=2.  270 

*o'~ 

=  21.92 

7ft. 

#b  =  6676.6 

Hc  =  295.65 

1.135 

TABLE  52o 
ORDINATES  FOR  MOMENT  INFLUENCE   LINES 


M4  Influence  Line, 

M  5  Influence  Line. 

Panel 

rim  =  rjom  —  [r)a  +  54  T/b  +7.9)}c] 

Tjm  =  rjom  —  [rja  +  45)?6  —  4.897  JJr] 

Point. 

ia 

'" 

/c 

ft, 

ft. 

ft. 

TjO 

54,6 

7.9,e 

rjm 

no 

45,b 

4.897>?c 

9* 

2 

8.2 

0.078 

0.100 

14.4 

4.21 

0.79 

1.20 

13.5 

3.51 

0.49 

2.  28 

4 

14.6 

0.104 

0.330 

28.8 

5.62 

2.61 

5.97 

27.0 

4.68 

1.62 

9.34 

6 

18.2 

0.088 

0.535 

25.2 

4.75 

4.23 

-1.98 

31.5 

3.96 

2.62 

11.96 

8 

20.0 

0.048 

0.647 

21.6 

2.59 

5.11 

-6.10 

27.0 

2.16 

3.17 

8.01 

10 

20.5 

0.000 

0.657 

18.0 

0.00 

5.19 

-7.69 

22.5 

0.00 

3.22 

5.22 

8' 

20.0 

-0.048 

0.647 

14.4 

-2.59 

5.11 

-8.12 

18.0 

-2.16 

3.17 

3.33 

6' 

18.2 

-0.088 

0.535 

10.8 

-4.75 

4.23 

-6.88 

13.5 

-3.96 

2.62 

1.8S 

4' 

14.6 

-0.104 

0  330 

7.2 

-5.62 

2.61 

-4.39 

9.0 

-4.68 

1.62 

0.70 

2' 

8.2 

-0.078 

0.100 

3.6 

-4.21 

0.79 

-1.18 

4.5 

-3.51 

0.49 

0.30 

All  ordinates  are  in  feet. 


ART.  52  SPECIAL  APPLICATIONS  OF  INFLUENCE  LINES  193 

The  three  influence  lines  for  the  redundants  are  now  drawn  by  constructing  the 
force  polygons,  using  the  wa,  wb  and  wc  forces  in  the  order  of  the  loaded  panel  points  and 
drawing  the  corresponding  equilibrium  polygons.  The  scales  used  for  the  w  forces  arc 
any  convenient  scales,  noting  that  the  pole  distance  must  of  course  be  laid  off  to  the  same 
scale  as  the  forces.  The  poles  Hb  and  Hc  were  divided  by  one  hundred,  thus  making 
the  tjb  and  rjc  ordinatcs  hundred  times  actual  when  measured  to  the  scale  of  lengths. 
The  wc  forces  being  of  both  signs  it  is  best  to  plat  their  algebraic  sums  from  some  initial 
point  of  the  load  line  and  number  the  points  in  the  order  of  the  loads,  thus  0,  1,2,  3,  etc., 
to  10.  The  pole  rays  are  drawn  in  the  same  order.  Note  the  check  by  which  the  end 
rays  of  the  Xa  line  intersect  in  a  point  on  the  y  axis. 

If  the  wa  forces  were  made  to  act  horizontally,  then  by  the  method  given  in  Art. 
38,  the  horizontal  resultant  of  these  forces  would  be  obtained  acting  at  the  intersection 
of  the  outer  rays.  This  resultant  EHwa  would  intersect  the  y  axis  in  the  center  of  gravity 
0.  However,  the  method  of  finding  zj  by  computation  is  preferable,  as  the  point  0 
must  be  accurately  located. 

The  moment  influence  line  for  any  point  as  panel  point  4,  is  now  constructed  by 
computing  the  i?w  ordinates  from  Eq.  (52o)  as  illustrated  in  Table  52o.  The  coordi- 
nates of  point  4  are  x  =54  ft.  and  y=7.9  ft.  and  the  rja,  T)b  and  rjc  ordinates  are  scaled 
from  the  three  X  influence  lines.  The  rj0  ordinates  are  obtained  from  the  Mo4  influence 
line  which  is  the  ordinary  moment  influence  line  for  point  4  of  a  simple  beam  of  span 
Z  =  180  ft.  The  table  indicates  the  details  of  combining  these  several  ordinates  to  obtain 
the  ordinates  which  are  finally  plotted  (Figs.  52c)  ,  giving  the  M4  influence  line.  The 
MS  line  is  similarly  found. 

The  M4  influence  line  may  be  used  to  obtain  the  stress  «S3_5  in  the  chord  3-5  with 
lever  r  =  15.21  ft.  Since  S=M/r,  this  same  influence  line  becomes  a  stress  influence 
line  with  a  factor  l/r. 

The  M5  influence  line  may  thus  be  regarded  as  the   stress  influence  line  for  the 
chord  4-6  with  a  factor  l/r  =  1  -*-  14.79  =0.068. 

Stress  influence  lines  might  also  be  developed  from  the  formula 


S  =S0  —SaXa  —SbXb  —SCX 


C, 


but  this  would  require  computing  the  stresses  Sb  and  Sc  in  addition  to  the  Sa  stresses 
already  found,  and  the  stresses  SA  and  SB  required  to  draw  the  ordinary  S0  line  for  any 
member  in  question. 

The  influence  lines  for  the  web  members  can  best  be  found  from  the  chord  influence 
lines  as  described  in  the  theoretical  portion  of  this  article. 

It  is  not  considered  necessary  to  go  further  into  this  problem,  as  the  theory  is  fully 
treated  and  the  general  method  is  illustrated  by  finding  a  single  stress  influence  line. 

The  temperature  stresses  are  determined  precisely  as  previously  outlined. 


CHAPTER  XI 
DESIGN   OF  STATICALLY  INDETERMINATE  STRUCTURES 

ART.  53.     METHODS    FOR   PRELIMINARY    DESIGNING 

The  term  indeterminate,  according  to  previous  definitions  given  in  Art.  2  and  the 
tests  for  identification  discussed  in  Art.  3,  is  always  used  in  the  sense  that  the  laws  ot 
pure  statics  fail  to  give  a  solution  when  applied  to  structures  involving  statically  inde- 
terminate or  redundant  conditions.  In  all  structures  of  this  class  the  analysis  of  stresses 
can  be  accomplished  only  by  resorting  to  the  theory  of  elasticity.  Hence,  indeterminate 
is  not  synonymous  with  impossible  only  in  so  far  as  the  application  of  the  laws  ot  statics 
is  concerned. 

The  complete  analysis  of  a  statically  indeterminate  structure,  according  to  Eqs 
(7 A)  and  (7n),  involves  the  solution  of  three  separate  problems  as  follows: 

1.  The  determination  of  all  internal  stresses  So,  resulting  in  the  principal  system 
from  the  externally  applied  loads  P. 

2.  The  determination  of  all  internal  stresses  St  and  Sr  produced  in  the  principal 
system  by  the  temperature  changes  and  abutment  displacements. 

3.  The  determination  of  elastic  deformations  in  the  principal  system  produced  by 
the  several  stresses  Sa,  Sb,  Sc,  etc.,  St  and  Sr  from  which  the  redundant  conditions  Xa, 
Xb,  Xc,  etc.,  Xt  and  Xr  may  be  evaluated. 

Eqs.  (7n)  and  (SD)  ,  furnish  the  solution  of  these  three  problems  for  any  indeterminate 
structure  by  means  of  considerations  which  are  of  similar  character  in  each  problem. 

A  statically  determinate  structure  can  always  be  analyzed  by  the  methods  known 
in  statics,  when  the  live  loads  and  the  dead  loads  of  the  structure  are  known.  Experience 
in  shop  practice  supplies  approximate  data  for  the  weights  of  structures  in  common  use, 
but  new  types  of  structures,  departing  from  the  ordinary  forms,  necessitate  repeated 
approximations  to  determine  the  dead  loads  before  a  final  analysis  for  the  given  live 
loads  becomes  possible. 

A  statically  indeterminate  structure  is  likewise  incapable  of  analyis  until  its  dead 
weight  is  known  with  some  degree  of  certainty  and  will  require  a  preliminary  design 
which  is  much  more  difficult  to  approximate  than  in  the  case  of  determinate  structures, 
because  this  involves  a  knowledge  of  the  magnitude  of  the  redundant  conditions.  These 
redundant  conditions  in  turn  require  that  the  cross-sections  of  the  members  be  also 
known. 

Fortunately,  the  influence  of  a  system  of  loads  can  be  made  to  depend  on  relative 
cross-sections,  thus  rendering  an  approximate  solution  of  the  redundants  X  possible, 

194 


ART.  53       DESIGN   OF  STATICALLY  INDETERMINATE   STRUCTURES  195 

before  the  actual  cross-sections  are  definitely  evaluated.  This  is  done  by  making  cer- 
tain assumptions  like  adopting  a  uniform  cross-section  for  all  chord  members  and 
neglecting  the  web  members  entirely,  whence  the  variable  F,  involved  in  p=l/EF  or 
Eq.  (42u)  ,  is  treated  as  a  constant.  The  w  loads  required  for  constructing  the  deflection 
polygon  for  the  loaded  chord  for  any  condition  X  =  l  may  then  be  made  EF  times  the 
values  given  by  Eq.  (36s)  ,  using  a  pole  distance  of  EF.  The  details  of  this  process  will 
be  discussed  later;  suffice  it  to  say  now  that  it  becomes  possible  to  construct  an  approx- 
imate influence  line  for  any  redundant  condition  without  knowing  the  final  cross-sections 
of  the  members. 

With  the  aid  of  these  approximate  X  influence  lines  the  redundants  may  be  evaluated 
for  any  given  live  loads  and  some  assumed  dead  loads.  Then  applying  all  these  loads 
simultaneously  to  the  principal  system,  the  first  approximate  values  for  the  stresses 
S  may  be  determined  from  a  Maxwell  .diagram.  The  redundants  X  ar*e  applied  along 
with  the  external  forces. 

These  stresses  S  will  now  serve  as  a  basis  for  a  close  approximation  of  the  cross- 
sections  F  from  which  our  new  influence  lines  for  the  redundants  X  may  be  found  in  the 
usual  manner  as  illustrated  by  the  problems  in  Chapter  X.  A  final  analysis  of  the  stresses 
is  now  possible  and  from  this  the  final  sections  are  found.  If  the  first  assumption  for 
dead  load  was  grossly  in  error  then  the  first  X  influence  lines  might  be  used  again,  employ- 
ing revised  dead  loads  before  the  final  analysis  is  made. 

This  process  is  on  the  whole  similar  to  the  one  above  cited  for  new  types  of  deter- 
minate structures,  the  dead  loads  for  which  are  not  accurately  known.  The  additional 
difficulty  here  encountered  consists  in  evaluating  the  redundants,  which  depend  both 
on  the  sections  and  loads  at  the  same  time.  Therefore,  all  such  work  is  likely  to  be  some- 
what tedious. 

The  method  of  determining  approximate  values  for  the  redundants  in  terms  of  the 
mutual  relations  of  the  sections  of  the  members  among  themselves,  instead  of  the  actual 
sections,  will  now  be  described. 

In  some  cases  as  for  arches  with  parallel  chords  it  is  always  permissible  to  assign 
a  constant  section  to  all  chord  members  and  neglect  the  web  system  entirely.  This 
is  then  a  very  simple  case  and  affords  a  ready  solution  for  the  X  influence  lines  by  com- 
puting all  w  loads  EF  times  too  large. 

When  there  is  only  one  redundant  condition  or  when  there  are  two  or  three  of 
these  so  chosen  as  to  comply  with  the  conditions  discussed  in  Art.  44  then  for  the  above- 
mentioned  case  of  constant  chord  sections  and  disregarding  the  web  system,  the  redundants 
may  be  found  analytically  from  Eq.  (42o)  whence 

S  ,53  , 

-      .      .      • 


where  the  stresses  S0  must  be  obtained  from  a  Maxwell  diagram  drawn  for  a  maximum 
case  of  combined  live  and  assumed  dead  loads  for  the  principal  system.     The  stresses 


196  KINETIC   THEORY  OF   ENGINEERING   STRUCTURES        CHAP.  XI 

Sa  are  found  for  the  conventional  loading  Xa  =  1  from  a  similar  stress  diagram,  and  these 
are  independent  of  any  actual  loads.  Eq.  (53  A)  shows  that  whenever  any  of  the  variables 
common  to  the  numerator  and  denominator  become  constant,  they  cease  to  affect  the 
value  of  Xa. 

The  stress  in  each  member  of  the  principal  system  may  now  be  found  from 

S=S0—SaXa,       ...     ........     (53s) 

and  the  first  approximate  sections  F  may  be  evaluated  from  these  stresses. 

The  approximate  effect  due  to  changes  in  temperature  may  also  be  included  b} 
finding  Xat  from  Eq.  (44c)  and  Eqs.  (SB)  and  (Sc)  as 

„       oat       ZSa£tl 

Aa<=^—  =  --  .—  ,       .........     (o3c; 

Oaa  * 


using  some  assumed  constant  value  for  F. 

If  for  any  reason  the  lengths  of  the  chord  members  are  all  equal,  then  I,  in  Eqs, 
(53A)  and  (53c)  ,  would  likewise  be  eliminated. 

A  somewhat  more  comprehensive  method,  especially  when  the  chord  sections  are 
not  assumed  equal,  is  obtained  by  drawing  the  influence  lines  for  the  redundants.  This 
method  will  now  be  outlined. 

When  the  redundants  are  so  chosen  that  the  simplified  Eqs.  (44s)  become  applicable 
then  the  X  influence  lines  may  be  approximated  by  computing  w  loads  for  relative 
sections  of  the  members.  Thus  for  any  redundant 


where  Fc  is  some  constant  chosen  about  equal  to  a  typical  chord  section  as  described 
below. 

The  wa  loads  from  Eqs.  (36s)  and  (36c),  for  condition  Xa  =  l,  then  become 


and  multiplying  each  by  EFC,  then, 

^.  .     (53E) 


ART.  53       DESIGN    OF  STATICALLY   INDETERMINATE  STRUCTURES  197 

Ma  is  the  moment,  with  lever  arm  r,  produced  by  the  conventional  loading  Za  =  l. 

If  a  deflection  polygon  be  drawn  for  these  EFcw  loads  with  pole  distance  unity,  the 
ordinates  r?  will  represent  values  EFcdma  measured  to  the  scale  of  lengths.  Hence  the 
numerator  of  Eq.  (53n)  represents  the  sum  of  the  products  Prj,  and  the  denominator  is 

2>S%pr  =  ^p/--jr=EFc'ZMaw,    according  to   Eq.   (53E),   thus    making  it    possible  to 

construct  any  X  influence  line  provided  suitable  assumptions  for  the  values  FC/F  can 
be  made. 

The  equation  of  the  X  influence  line  for  a  load  P  =  1  then  becomes 

77T  TJT     ^  77T  TJ1    JJt 

Zj2jrcOma          Hircuma 
„  —  — -  = — . f^F^ 

7^       FW^  1\/T  iii  *  •      ••«••      ^uor^ 

2>S~l-pT 

in  which  Fc  need  not  be  numerically  known,  but  the  ratios  FC/F,  for  different  members 
of  the  frame,  must  be  approximated  in  order  to  compute  the  EFcw  loads  from  Eqs. 
(53E). 

The  value  of  Fc  should  be  chosen  equal  to  that  of  a  chord  section  of  most  frequent 
occurrence,  which  would  usually  be  nearest  the  crown  of  an  arch  or  where  the  chord 
approaches  the  horizontal.  This  will  make  Fc/F  =  l  for  most  chord  members  and 
usually  that  assumption  may  be  made  for  both  top  and  bottom  chord  members.  The 
web  system  may  be  entirely  neglected  in  the  first  approximation  or  if  it  is  thought 
desirable  to  include  the  web  members  then  FC/F  may  be  given  some  constant  value 
ranging  from  2  to  10  for  these  members,  depending  on  the  character  of  the  structure. 

The  Fc  should  not  be  regarded  as  an  average  value  of  the  chord  sections  because 
the  deflection  polygon  of  a  frame  is  governed  largely  by  the  chords  near  the  center  of  the 
span.  Thus,  if  the  depth  of  an  arch  is  much  greater  at  the  springing  than  at  the  crown, 
then  the  crown  sections  will  govern. 

From  the  approximate  X  influence  lines  the  redundant  conditions  may  be  evaluated 
for  any  case  of  total  loading  and  the  stresses  S0,  Sa,  etc.,  can  be  found  by  Maxwell  dia- 
grams or  otherwise.  Then  Eq.  (53B)  will  furnish  the  stresses  S  from  which  the  sections 
of  the  several  members  are  derived.  The  stresses  S  may  also  be  obtained  from  a  single 
stress  diagram  by  applying  all  redundants  and  external  loads  simultaneously  to  the  prin- 
cipal system.  Temperature  effect  may  be  included  in  the  redundants  at  the  same  time. 

Another  method  of  making  a  preliminary  design  would  result  from  the  use  of  a 
Williot-Mohr  displacement  diagram  as  used  in  Art.  50.  However,  the  deflection  poly- 
gons as  found  from  the  w  loads  are  more  easily  obtained  except  for  oblique  loads  P. 

To  illustrate  the  entire  procedure  of  making  a  preliminary  design  for  a  structure 
involving  redundant  conditions,  the  fixed  arch  in  Art.  52  will  now  be  investigated. 

Example.  The  data  for  this  arch  are  given  in  the  example  of  Art.  52  and  need  not  be 
repeated  here. 

The  EFcwa  loads  will  be  computed  from  Eq.  (53E)  making  FC/F  =  1  and  neglecting 
the  web  system.  The  moment  Ma  =  I  and  the  lengths  I,  lever  arms  r  and  ordinates  z 
are  taken  from  Tables  52A  and  52c. 


198 


KINETIC  THEORY  OF  ENGINEERING  STRUCTURES          CHAP.  XI 


TABLE  53A 
COMPUTATION   OF   APPROXIMATE   LOADS   EFcwc 


Pin 

Point. 

Chord. 

I 
Feet. 

r 
Feet. 

1 

~2 

Feet. 

z 
Feet, 

EFcwa^~ 

zEFcwn 

1 

0-2 

20.73 

14.92 

0  .  00449 

0.00 

0.0931 

0.000 

3 

2-4 

19.51 

14.84 

0.00454 

9.99 

0.0886 

0.885 

5 

4-6 

18.72 

14.79 

0.00457 

17.03 

0.0856 

1.458 

7 

6-8 

18.25 

14.77 

0.00458 

21.53 

0.0836 

1.800 

9 

8-10 

18.03 

14.79 

0.00457 

23  .  69 

0.0824 

1.952 

2 

1-3 

20.59 

14.14 

0.00500 

22.31 

0.1030 

2.300 

4 

3-5 

19.33 

15.21 

0.00432 

29  .  83 

0.0835 

2.491 

6 

5-7 

18.55 

15.25 

0.00430 

34  .  99 

0  .  0798 

2.792 

8 

7-9 

18.33 

15.28 

0.00428 

38.01 

0  .  0784 

2.980 

10 

4(9-9') 

9.00 

15.31 

0.00427 

39.00               0.0384 

1.498 

0.8164 

18.156 

This  makes  z0'  = 


2zEFcwa      18.156 


=  22.24  ft.  from  which  the  origin  0  can  be  located 


ZEFcwa      0.8164 

and  the  ordinates  y=z—z0'  be  computed.     The  remaining  w  loads  and  pole  distances 
are  then  easily  found. 

TABLE  53 B 
APPROXIMATE   w  LOADS  AND   POLE   DISTANCES 


Coordinates. 

Pin 

Point. 

EFcWa 

X 

y 

EFcwb  = 

xEFcWa 

EFcwc  = 
yEFcwa 

xEFcWb 

yEFcWc 

Feet. 

Feet. 

1 

0.0931 

81 

-22.24 

7.541 

-2.070 

610.82 

46:  04 

2 

0.1030 

72 

+   0.07 

7.416 

+  0.007 

533.95 

0.00 

3 

0.0886 

63 

-12.25 

5.582 

-1.085 

351.67 

13.29 

4 

0.0835 

54 

+   7.59 

4.509 

+  0.634 

243.49 

4.81 

5 

0.0856 

45 

-   5.21 

3.852 

-0.446 

173.34 

2.32 

6 

0.0798 

36 

+  12.75 

2.873 

+  1.017 

103.43 

12.97 

7 

0.0836 

27 

-  0.71 

2.257 

-0.059 

60.94 

0.04 

8 

0.0784 

18 

+  15.77 

1.411 

+  1.236 

25.40 

19.49 

9 

0.0824 

9 

+    1.45 

0.742 

+  0.120 

6.68 

0.17 

4(10) 

0.0384 

0 

+  16.76 

0.000 

+  0.644 

0.00 

10.79 

Totals.. 

0.8164 

36  183 

—  3  660 

2109  72 

109  92 

Ha=  1.633 

+  3.658 

Hb  =  4219.  4 

tfc  =  219.8 

The  three  influence  lines  for  Xa,  Xb,  and  Xc  may  now  be  drawn  precisely  as  was  done 
in  Figs.  52c,  by  using  the  values  in  Table  53B. 


ART.  53        DESIGN  OF  STATICALLY  INDETERMINATE  STRUCTURES  19i) 

Comparing  the  ordinates  found  from  these  influence  lines,  Fig.  53A,  with  those  pre- 
viously obtained  in  Art.  52,  it  will  be  seen  that  the  former  are  fairly  close  to  the  correct 
values,  but  inclined  to  be  a  little  large. 


X,,  INFLUENCE  LINE  -  ORDtNATES  ACTUAtTOSCALEOF  LENGTHS- 


ISO' 


FlG.  53A. 

The  live  load  per  truss  per  panel  was  stipulated  at  30  kips  and  for  an  assumed  dead 
load  of  20  kips  the  total  load  would  be  50  kips.  . 

Using  the  ordinates  from  the  influence  lines  in  Fig.  53A  and  the  total  load  of  50  kips 
per  truss  per  panel,  the  redundants  Xa,  Xb  and  Xc  are  computed  in  the  following  Table 
53c.  For  a  symmetric  loading  Xb=0  and  need  not  be  considered. 


200 


KINETIC  THEORY  OF  ENGINEERING  STRUCTURES 


CHAP.  XI 


TABLE  53c 
REDUNDANTS   A'a  AND   Xc  FOR  TOTAL   LOADING 


Ordinates. 

Redundants. 

Panel 

Panel 

Point. 

• 

f]a 

TjC 

Xa 

Xc 

Feet. 

Feet. 

Kips. 

Kip-  feet. 

Kips. 

2 

8.4 

0.086 

50 

420 

4.3 

Xa  =  2X3730  =  7460     k.  ft. 

4 

14.5 

0.298 

50 

725 

14.9, 

Xc  =  2X97.9  =    195.8  kips 

6 

19.0 

0.520 

50 

950 

26.0 

A0  =  iSF  =  225  kips 

8 

21.5 

0.684 

50 

1075 

34.2 

i(10) 

22.4 

0.740 

25 

560 

18.5 

Totals 

225 

3730 

97.9 

The  redundants  might  be  obtained  for  any  position  of  the  live  load,  but  since  the 
chords  are  stressed  to  their  maximum  for  total  loading  over  the  span,  and  since  the  web 
system  plays  a  rather  unimportant  part  in  structures  of  the  class  here  considered,  no 
further  investigation  of  stresses  is  warranted  at  this  point. 

The  stresses  for  the  total  loading,  including  the  redundants  Xa  and  Xc  applied  as 
external  forces,  are  now  found  from  a  Maxwell  stress  diagram  in  Fig.  53 B,  and  from  these 
the  preliminary  cross-sections  of  the  members  are  deduced. 

From  the  sums  in  Table  53A,  the  value  of  z</=22.24  ft.  This  locates  the  pole  0 
on  the  y  axis.  The_rigid  disk,  upon  which  the  redundants  act,  may  now  be  replaced  by 
two  rigid  members  10  and  00.  For  the  symmetric  total  loading,  Xb=0,  and  hence  the 
external  loads  are  Xc  applied  at  0;  and  a  moment  Xa  applied  anywhere  to  the  disk;  a 
vertical  end  reaction  A0;  and  the  loads  P  acting  at  the  several  top  chord  panel  points. 

In  order  that  the  moment  Xa  may  be  incorporated  in  the  stress  diagram,  it  must  be 
resolved  into  two  equal  and  opposite  forces  acting  on  the  rigid  disk.  Since  these  forces 
may  be  applied  anywhere  on  the  disk,  it  is  best  to  choose  the  lever  arm  1/2,  thus  making 
the  forces  V  =2Xa/l  each,  one  acting  upward  at  the  support  A  and  the  other  applied 
downward  at  the  pole  0.  The  forces  V  and  Xc,  acting  at  0,  are  then  combined Jnto  a 
resultant  Z  and  from  this  resultant  the  stresses  in  the  rigid  members  10  and  00  are 
found,  making  it  possible  now  to  draw  the  stress  diagram  beginning  at  the  reaction  A. 
The  remainder  of  the  work  offers  no  particular  difficulty. 

It  might  be  desirable  to  test  out  the  resultant  polygon  with  the  aid  of  Eqs.  (52s). 
From  these  20=Za/Xc=38.10  ft.  and  the  ordinate  CD,  Fig.  53s,  determines  the  inter- 
section of  the  resultants  R,  RI  and  R2  in  the  point  C.  The  closing  line  aibj.  for  symmetric 
loading,  is  fully  located  when  OD=z0  is  given,  hence  the  points  a}  and  61  are  found  on 
the  horizontal  through  D.  The  resultant  RI  is  obtained  by  combining  A0  and  Xe  at  a! 
and  R2  is  the  resultant  of  B0  and  Xe  at  6t.  The  resultant  R  =  2P  must  be  in  equilibrium 
with  RI  and  R2  (see  Fig.  52A) ,  hence  the  three  forces  must  intersect  in  a  point  C. 

A  further  check  is  obtained  by  computing  the  ordinate  CI>=Z#/4XC  =  103.4  ft. 

The  stresses  S  are  now  tabulated  and  cross-sections  F  are  determined  in  Table  53D, 
using  a  rather  low  unit  stress  of  say  9200  Ibs.  per  sq.in.,  instead  of  an  allowable  15,000 


ART.  53       DESIGN  OF  STATICALLY  INDETERMINATE  STRUCTURES  201 

Ibs.     This  then  covers  extra  metal  for  joints,  splices,  latticing,  etc.,  and  for  reduction 
to  net  sections  for  tension  members,  and  for  l/r  for  columns. 


o       BOO  KIM. 


FIG.  53s. 

TABLE  53D 
PRELIMINARY  STRESSES  AND   SECTIONS 


Top  Chord. 

Bottom  Chord. 

Diagonals. 

Diagonals. 

Mem. 

8 

Kips. 

F 

Sq.in. 

Mem. 

S 
Kips. 

F 
Sq.in. 

Mem. 

S 
Kips. 

F 
Sq.in. 

Mem. 

S 
Kips. 

F 

Sq.in. 

0-2 
2-4 
4-6 
6-8 
8-10 

-  920 
-  1028 
-1114 
-1178 
-1210 

100 
110 
120 
124 
130 

1-3 
3-5 
5-7 

7-9 
9-9' 

748 
846 
930 
970 
988 

82 

94 
100 
104 
128 

0-1 
2-3 
4-5 

6-7 
8-9 

-  70 

+  166 
+  160 
+  117 
+   95 

8 
18 
18 
14 
10 

1-2 
3-1 
5-6 

7-8 
9-10 

140 
-  90 
-   55 

+  10 

+  48 

14 
10 
6 
6 
6 

202  KINETIC   THEORY  OF  ENGINEERING  STRUCTURES          CHAP.  XI 

These  cross-sections  were  used  in  the  example  of  Art.  o2,  in  making  the  complete 
analysis. 

Naturally  a  two-hinged  arch  or  other  structure  involving  one  redundant  only, 
would  be  susceptible  to  the  same  method  of  design  above  outlined,  though  the  labor  would 
be  greatly  lessened. 

Foi  an  unsymmetric  arch  the  angle  /?  would  have  to  be  determined- and  the  closing 
line  aibi  would  not  be  horizontal. 

ART.  54.     ON   THE    CHOICE   OF   THE    REDUNDANT   CONDITIONS 

In  the  analysis  of  a  structure  involving  redundancy,  it  becomes  necessary  to  remove 
the  redundant  conditions,  whether  external  or  internal,  and  thereby  reduce  the  indeter- 
minate system  to  a  determinte  principal  system  to  which  the  redundant  conditions  are 
applied  along  with  the  external  loading. 

The  particular  reactions  or  members  best  suited  to  represent  the  redundant  con- 
ditions designated  by  Xa,  Xj,,  Xc,  etc.,  are  those  which  reduce  the  given  structure  to  the 
simplest  possible  principal  system.  This  will  usually  offer  no  difficulty. 

However,  many  cases  present  themselves  wherein  it  would  be  difficult  to  decide 
which  of  several  possible  assumptions  would  represent  the  most  judicious  selection  of  the 
redundants  in  the  direction  of  simplifying  the  analysis.  Therefore,  a  few  suggestions 
along  these  lines  will  not  be  out  of  place. 

According  to  the  general  method  just  referred  to,  the  elastic  deformation  of  an 
indeterminate  structure  subjected  to  loads  P,  will  be  exactly  the  same  as  that  of  its  prin- 
cipal system  loaded  with  loads  P,  Xa,  Xb,  Xc,  etc.  In  each  case  the  deformation  is 
entirely  derived  from  the  elastic  changes  Al  in  the  lengths  of  the  members  of  the  principal 
system. 

The  structural  deformation  is  thus  affected  by  the  redundant  conditions  merely 
to  the  extent  of  altering  the  stresses  in  the  necessary  members  of  the  principal  system. 

Hence,  for  the  same  case  of  loading,  the  stresses  S0  in  the  principal  system  are 
always  diminished  by  the  redundant  conditions  provided  temperature  stresses  and 
abutment  displacements  are  excluded. 

Generally  speaking,  any  member  or  reaction  of  an  indeterminate  structure  may  be 
removed  to  produce  the  principal  system  so  long  as  the  latter  still  remains  a  stable, 
determinate  structure  and  does  not  become  subject  to  infinitesimally  small  rotation, 
a  condition  described  in  Art.  3,  Fig.  3c. 

The  rule  should  be  to  select  a  principal  system  of  the  simplest  possible  form,  always 
avoiding  composite  structures,  such  as  the  three-hinged  arch  or  a  cantilever,  whenever 
a  simple  beam  or  truss  could  as  well  be  used. 

Solid  web  structures  should  always  be  transformed  into  externally  determinate  beams 
by  assigning  the  redundant  conditions  to  the  supports. 

Framed  structures,  if  externally  indeterminate,  should  always  be  so  transformed 
as  to  remove  the  external  conditions.  The  only  exception  to  this  rule  might  be  a  con- 
tinuous girder  wherein  a  top  chord  member  over  each  intermediate  pier  might  be  treated 
as  a  redundant  member. 


ART.  55        DESIGN  OF  STATICALLY  INDETERMINATE  STRUCTURES  203 

In  the  case  of  the  fixed  arch,  always  involving  three  redimdants,  several  assumptions 
may  be  made.  First  removing  one  fixed  end  and  thus  reducing  the  arch  to  a  simple 
cantilever  arm  acted  upon  by  a  moment  Xa,  a  vertical  force  Xb  and  a  horizontal  force  Xc. 
Second,  by  removing  two  members  adjacent  to  one  end  support  and  one  member  adjacent 
to  the  other  end  support,  thus  forming  a  girder  on  two  determinate  supports  acted  upon 
by  three  external  forces  replacing  the  three  members  thus  removed. 

Also  three  members  may  be  removed  at  the  crown  converting  the  arch  into  two 
cantilever  arms  with  three  external  forces  Xa,  Xb,  and  Xc  applied  to  the  end  of  each  arm 
to  replace  the  redundant  members. 

Frequently  all  the  redundant  conditions  (not  exceeding  three)  may  be  applied  at 
one  point  as  described  in  Art.  44.  Then  for  certain  assumed  directions  of  the  X's,  the 
work  equations  become  exceedingly  simple  and  afford  excellent  graphic  solutions.  The 
best  of  these  possible  solutions  was  chosen  in  Art.  52,  where  a  complete  fixed  arch  problem 
is  solved. 

Indeterminate  structures,  having  a  vertical  axis  of  symmetry,  will  always  have  two 
d's  bearing  double  subscripts  of  like  letters,  which  become  equal.  Thus  daa=dbb,  which 
condition  greatly  simplifies  the  determination  of  Xa  and  Xb,  as  illustrated  in  the  case  of 
a  girder  over  four  supports  with  the  two  outer  spans  equal.  See  Fig.  43A.  In  this  and 
similar  problems,  when  the  above  mentioned  symmetry  exists,  the  two  influence  areas 
for  Xa  and  Xb  will  also  be  equivalent  but  symmetrically  placed. 

When  the  redundant  conditions  are  internal  then  the  only  way  of  deriving  the  princi- 
pal system  is  to  remove  such  redundant  members  and  replace  them  by  external  forces  A'. 

Composite  structures,  or  those  composed  of  several  determinate  frames  combined 
into  an  indeterminate  system,  are  best  treated  by  assigning  the  redundant  conditions 
to  the  reactions  and  treat  as  for  external  redundancy. 

ART.  55.     SOLUTION   OF   THE   GENERAL   CASE    OF   REDUNDANCY 

In  the  following  it  will  be  assumed  that  the  preliminary  design  is  completed  and  it 
is  now  desired  to  make  the  final  analysis  for  stresses  due  to  any  causes  such  as  loads, 
temperature  and  abutment  displacements. 

Problems  of  the  general  type  are  usually  quite  complicated,  and  since  it  is  important 
to  know  the  stresses  due  to  loads,  to  temperature  changes  and  to  yielding  supports  sepa- 
rately, these  should  always  be  dealt  with  in  this  manner.  This  is  not  done  merely  as  a 
matter  of  convenience,  but  it  is  necessary  to  know  the  relation  between  the  load  and 
temperature  stresses  as  a  criterion  in  judging  the  merits  of  any  particular  structure 
under  consideration. 

In  the  general  discussion  which  follows  here,  only  the  load  effects  will  be  included, 
while  the  other  matters  will  be  taken  up  later.  Hence  the  quantities  2#Jr  and  dt,  St, 
etc.,  will  all  be  neglected,  assuming  for  the  present  that  they  are  all  zero. 

To  make  the  case  perfectly  general,  no  distinction  will  be  made  between  external 
and  internal  redundancy,  hence  any  of  the  redundants  may  be  assumed  to  belong  to 
either  class. 

For  n  redundant  conditions,  Eqs.  (?A),  offer  a  general  solution  for  the  stress  S  in  any 


204 


KINETIC   THEORY  OF   ENGINEERING  STRUCTURES 


CHAP.  XI 


member,  the  moment  M  about  any  point,  the  shear  Q  on  any  section  or  any  of  the 
reactions  R,  all  for  the  principal  system  only.  The  redundant  conditions  X,  involved 
in  these  equations,  are  treated  as  external  forces  applied  to  the  principal  system  along 
with  the  loads  P  and  their  reactions.  These  redundants  may  be  found  from  Mohr's 
work  equations  (?H)  or  (80)  of  which  there  will  always  be  as  many  equations  as  there  are 
redundants,  rendering  the  solution  of  the  X's  possible  by  solving  for  simultaneous  values 
of  n  unknowns  X  in  n  equations. 

It  is  usually  perferable  to  employ  Eqs.  (80)  in  terms  of  deflections  which  are 
obtained  from  deflection  polygons  of  the  loaded  chord,  though  in  principle  the  solution 
remains  the  same  whether  dealing  with  conventional  stresses  or  conventional  deflections. 

In  general  then  for  n  redundants,  including  effects  due  to  temperature  changes  and 
yielding  supports,  Eqs.  (?A)  give 


(55A) 


S=S0   —  SaXa  —  SfrXb  —  etc.  —  SnXn 
M=M0-MaXa-MbXb-  etc.  -MnXn±Mt+Mr 
R=R0  -  RaXa- RbXb-  etc.  -RnXn±Rt  +R 
Q=*Qo  -  QaXa-  QbXb-  etc.   -  QnXn±Qt  +Q 


wherein  tne  quantities  S0,  M0,  R0  and  Q0  are  all  linear  functions  of  the  externally  applied 
loads  P  acting  on  the  principal  system  as  a  result  of  what  is  known  as  condition  .X"=-0. 
The  quantities  bearing  subscripts  a,  b,  c,  etc.,  to  n,  are  constants  due  to  conventional 
loadings  Xa  =  l,  Xb  =  l,  Xc  =  l,  etc.,  to  Xn  =  l,  while  the  subscript  prefers  to  temperature 
effects  and  the  subscript  r  to  effects  produced  by  yielding  supports. 

The  n  redundants  may  be  obtained  from  Eqs.  (7n)  in  terms  of  the  constant  stresses 
due  to  the  conventional  loadings  or  from  Eqs.  (So)  in  terms  of  certain  deflection  constants 
obtained  from  the  same  conventional  loadings.  The  latter  equations  are 


dn  = 


B  —Xadna  — 


etc.  —X^dan  — 

etc.  —Xndbn  — 

etc.  —  Xndcn  — 

etc.  —Xndnn  — 


+dc 


(55B) 


wherein  the  o's  have  the  definitions  given  in  Art.  8,  and  all  those  bearing  double  sub- 
scripts of  like  letters  are  equal  by  Maxwell's  law.     Thus 


=dc 


=d      etc.  d    = 


etc. 


meaning  that  the  order  of  the  subscripts  is  immaterial  and  that  only  half  of   the  d's 
need  be  determined,  or  if  they  are  all  determined,  that  they  must  check. 

Hence,  either  one  of  the  subscripts  may  be  made  to  refer  to  the  point  of  application 
of  a  load,  while  the  other  subscript  deals  with  the  conventional  loading  or  condition 
X  —  \.  The  terms  RAr  express  the  effect  due  to  yielding  supports  and  the  dt  quantities 


ART.  55        DESIGN  OF  STATICALLY  INDETERMINATE  STRUCTURES  205 

express  the  effect  of  temperature  changes,  both  of  which  will  be  placed  equal  to  zero 
for  the  present,  though  they  will  be  considered  in  following  articles. 

The  redundants  X  may  represent  indeterminate  reaction  conditions  or  they  may 
be  stresses  in  redundant  members,  and  in  either  case  the  displacements  da,  db,  Sc,  etc., 
on  are  the  displacements  of  the  points  of  application  of  the  respective  redundants  in  the 
directions  of  their  lines  of  action.  These  values  o  may  thus  be  expressed  in  terms  of 
the  redundants  themselves  in  accordance  with  Eqs.  (7j).  When  the  point  of  application 
of  a  certain  redundant  is  rigid,  then  its  d  =0. 

For  each  of  the  n  redundant  conditions  there  will  be  one  work  equation  of  the  form 
of  Eqs.  (55B)  involving  deflections  due  to  conventional  loadings.  These  together  with 
the  case  of  actual  loading  due  to  loads  P,  will  constitute  in  all  n  +  l  cases  of  loading  on 
the  principal  system  to  solve  for  one  position  of  a  moving  train  of  loads. 

The  aim,  in  all  practical  problems  of  this  nature,  consists  in  representing  all  the 
required  stresses  or  other  functions  by  influence  lines,  thus  requiring  n  -f  1  influence  line-* 
to  determine  each  such  stress  or  function.  However,  the  n  influence  lines  for  the  n  redun- 
dants will  be  the  same  for  all  cases  and  all  stresses  or  functions  of  the  same  structure, 
while  the  influence  line  for  the  load  effect  must  be  separately  found  for  each  stress, 
moment,  shear  or  reaction. 

For  influence  lines  the  applied  load  becomes  unity  and  the  functions  ^Pmdm  =  1  •  dm. 
Each  of  the  n  redundants  will  furnish  a  deflection  polygon  for  condition  X  =  l,  drawn 
for  the  loaded  chord  of  the  principal  system.  The  n  deflection  polygons  will  then 
furnish  all  the  conventional  deflections  in  Eqs.  (55s)  for  a  load  P  =  l.  Also,  since  the 
subscripts  may  be  interchanged,  one  such  deflection  polgyon  drawn  for  XA  =  l  will  fur- 
nish all  the  double  subscript  bearing  o's  of  the  first  equation. 

Thus  as  a  matter  of  convenience,  all  of  the  d  coefficients  in  a  single  equation 
can  be  determined  from  one  deflection  polygon  drawn  for  the  loaded  chord  of  the  prin- 
cipal system.  The  same  might  also  be  accomplished  by  drawing  a  Williot-Mohr  dis- 
placement diagram  for  the  principal  system,  though  this  would  usually  be  more  laborious. 

Therefore,  the  n  Eqs.  (55s)  can  be  solved  successively  with  the  aid  of  n  deflection 
polygons,  in  accordance  with  the  following  form  with  transposed  subscripts  and  for  a 

single  load  P  =  l. 

X  I 


•y-  7 


X2 

XbOhc-Xcdcc       .    .    .     -X^nc  = 


,    .     .     .     (5oc) 


where  the  values  of  da,  3b,  Sc,  etc.,  were  obtained  from  Eqs.  (7j)  and  may  be  chosen  to 
represent  anv  of  the  possible  cases  of  redundancy,  as  changes  in  the  lengths  of  redundant 
members  elastic  displacements  of  redundant  supports  or  angular  changes  between  pairs 


206  KINETIC  THEORY  OF  ENGINEERING  STRUCTURES          CHAP.  XI 

of  lines  according  to  Art.  9.  When  the  redundants  are  immovable  reactions,  then  these 
S's  become  zero. 

The  d  coefficients  in  Eqs.  (55c)  can  thus  be  obtained  from  n  deflection  polygons 
drawn  as  above  described  and  the  n  equations  solved  for  simultaneous  values  of  the  X's 
thus  furnish  the  values  for  a  complete  solution  of  Eqs.  (OOA). 

Eqs.  (55c)  are  written  for  a  single  load  P  =  l  intended  specifically  for  use  with  influence 
lines,  which  are  most  valuable  when  dealing  with  concentrated  load  systems. 

Examples  involving  dead  loads  only,  or  when  the  live  load  cannot  occupy  more  than 
one  or  two  positions,  could  be  solved  advantageously  without  resorting  to  influence  lines 
according  to  the  very  interesting  example  of  a  lock  gate  in  Chapter  XIV. 

For  solid  web  structures  and  masonry  arches,  all  the  above  is  applicable,  remember- 
ing that  in  these  cases  the  redundants  must  all  be  external.  As  soon  as  the  X's  are  found 
the  stresses  on  any  section  are  readily  ascertained. 

The  solution  of  the  n  simultaneous  equations  is  a  matter  of  considerable  labor  and 
may  offer  some  difficulty.  The  method  best  suited  to  problems  of  the  kind  here  con- 
sidered is  given  in  connection  with  the  lock  gate  problem  referred  to  above.  See  Chap- 
ter XIV,  Art.  68. 

ART.  66.     EFFECT   OF   TEMPERATURE    ON   INDETERMINATE   STRUCTURES 

Determinate  systems  are  not  materially  affected  by  temperature  changes.  But  it 
is  not  correct  to  say  that  the  temperature  stresses  are  zero,  because  any  structure  which, 
in  consequence  of  changes  in  temperature,  undergoes  such  deformations  as  to  change 
the  positions  of  the  points  of  application  of  the  external  forces  must  thereby  create  some 
stresses.  These  will  usually  be  small  and  are  entirely  negligible  except  possibly  in  very 
flat  suspension  systems  or  very  flat  three-hinged  arches  where  the  horizontal  pull  or  thrust 
is  a  function  of  the  middle  ordinate. 

For  the  case  of  indeterminate  structures,  wherein  the  redundant  conditions  ares 
direct  functions  of  elastic  deformations,  the  general  rule  implies  that  stresses  always 
accompany  changes  in  the  lengths  of  structural  dimensions,  no  matter  what  external 
cause  may  have  produced  such  changes. 

All  temperature  changes  in  externally  indeterminate  structures  will  affect  the  reac- 
tions and  these  in  turn  set  up  temperature  stresses  in  the  members.  The  only  exception 
to  this  case  is  a  continuous  girder  with  supports  on  the  same  level  and  subjected  to  abso- 
lutely uniform  temperature  throughout.  When  such  a  structure  is  unequally  heated 
then  excessive  temperature  stresses  may  result,  as  shown  in  the  examples  of  Arts.  47; 
and  48. 

On  the  other  hand  where  the  redundancy  is  entirely  internal,  no  temperature  stresses 
are  produced  so  long  as  the  whole  structure  retains  a  uniform  temperature  throughout. 
When  this  uniformity  does  not  exist,  then  temperature  stresses  are  created,  though  the 
reactions  may  or  may  not  be  materially  affected. 

Hence,  in  all  cases  of  redundancy,  it  will  be  necessary  to  investigate  the  question 
of  temperature  stresses  in  a  very  thorough  manner,  as  in  many  instances  these  may 
assume  dangerous  proportions. 


ART.  so        DESIGN  OF   STATICALLY  INDETERMINATE  STRUCTURES  207 

For  this  reason  also,  statically  determinate  structures  should  always  receive  the 
preference,  other  considerations  being  nearly  equal. 

Internal  redundancy  is  less  objectionable  than  external,  and  according  to  the  best 
modern  engineering  experience,  no  design  should  be  made  to  include  more  than  one 
redundant  condition  of  any  kind.  This  is  especially  true  of  steel  structures,  though  to 
a  lesser  degree  applicable  to  masonry  arches  of  short  spans  where  the  poor  conductivity 
of  masonry  acts  as  a  protective  agency  against  excessive  temperature  deformations. 

For  long-span  masonry  arches  this  consideration  assumes  greater  importance.  Even 
though  monumental  structures  of  this  class  have  been  built  and  are  regarded  with  pride 
and  admiration  from  an  esthetic  standpoint,  they  cannot  be  accepted  as  representative 
of  the  best  practice  when  viewed  from  the  point  of  modern  and  progressive  engineering. 

The  general  effect  of  temperature  changes  on  indeterminate  systems  is  thus  to  pro- 
duce deformations  and  resultant  stresses  of  the  same  kind  as  those  created  by  externally 
applied  loads  in  accordance  with  Eqs.  (55A). 

Since  the  temperature  effects  may  be  plus  or  minus  in  character,  depending  on  the 
existing  conditions,  it  is  always  possible  to  increase  the  stresses  due  to  the  loading  by 
choosing  appropriate  changes  in  temperature.  For  this  reason  the  temperature  effect 
is  applied  with  a  plus  sign  in  Eqs.  (55A),  meaning  thereby  that  the  function,  whether  posi- 
tive or  negative,  suffers  a  numerical  increase.  Of  course  some  assumptions  will  produce 
a  decrease,  but  these  are  not  vital  to  the  problem  and  require  no  consideration. 

It  is  always  desirable  to  determine  the  temperature  stresses  apart  from  all  other 
influences  so  as  to  furnish  a  clear  conception  of  their  relative  importance  to  the  load 
stresses.  This  then  offers  a  criterion  by  which  to  judge  the  feasibility  of  a  structure 
involving  redundancy. 

The  temperature  stresses  are  determined  from  Eqs.  (7n)  and  (80)  by  dropping  out 
all  terms  depending  on  Pm,  S0  and  R,  thus  reducing  these  equations  to  the  following 
forms,  covering  all  members  of  the  principal  system: 

da=eaa  =  ^Sastt-Xat^S^p   -Xbt2SaSbp  .  .  .  etc.  1 

k       ...     (56A) 
db  =  eilb  =  '2Sbetl-Xat2SJSbp-XbtZ,Szp     .  .  .  etc.  J 

or  from  Eqs.  (80) 

8a=Xatpa=8at-Xatdaa-Xbt8ab    •    •    •    etc.    ] 

\ (o6B) 

8b=Xbtpb=dbt-Xat8btt-Xbt8bb   ...  etc.  J 

In  both  Eqs.  (56A)  and  (56s)  the  displacements  da  and  db  become  zero  for  external 
redundancy  with  immovable  supports. 

Accordingly  the  redundant  conditions  Xat,  Xbt,  etc.,  may  be  found  from  either  set  of 
the  above  equations,  depending  on  the  method  used  in  finding  the  load  effects  from  Eqs. 
(7n),  or  (80)  since  both  these  and  Eqs.  (56A)  and  (56u)  involve  the  same  constants. 

The  stresses  St  are  then  easily  found  from  a  Maxwell  stress  diagram  drawn  for  the 
principal  system  with  the  forces  Xat,  Xbt,  etc.,  applied  as  external  forces.  See  also  examples 
in  Arts.  47,  48,  49  and  50. 


208  KINETIC  THEORY  OF  ENGINEERING  STRUCTURES          CHAP.  XI 

Regarding  the  assumptions  which  may  be  made  as  a  basis  for  computing  temperature 
stresses,  the  following  conclusions  are  taken  from  some  experiments  on  steel  arches  at 
Lyons,  France,  and  given  in  An.  d.  ponts  et  chaus.,  1893,  II,  p.  438-444. 

For  an  air  temperature  in  the  sun  of  5°  F.  higher  than  the  shade  temperature,  when 
the  latter  was  90°  F.,  the  steel  had  an  average  temperature  of  115°  F.,  while  the  parts 
exposed  directly  to  the  sun  were  heated  to  about  130°  F.  and  the  shaded  portions  indicated 
about  104°  F.  At  the  same  place  the  coldest  winter  temperature  was  about  —15°  F. 

The  steel  was  thus  subjected  to  a  mean  range  of  — 15°  to  +115°  =  130°  F.  giving 
an  average  of  +65°  F.  Therefore,  such  a  structure  should  be  designed  for  a  mean 
temperature  of  65°  F.,  allowing  for  uniform  changes  of  ±65°  F.  from  this  mean. 

The  difference  between  maximum  and  minimum  simultaneous  temperatures  in 
the  steel,  amounting  to  26°  F.  in  these  experiments,  may  cause  very  serious  stresses  in 
certain  structures  like  fixed  arches  and  continuous  girders  over  several  supports.  In 
the  latter  case  the  entire  top  chord  would  elongate  relatively  to  the  bottom  chord  and 
thus  set  up  an  arching  effect,  relieving  the  intermediate  supports  and  increasing  the 
two  end  supports,  while  a  uniform  change  in  temperature  would  produce  no  stresses. 

Structures  over  a  single  span  would  not  be  so  seriously  stressed  for  unequal  temper- 
atures in  the  two  chords.  In  fact  this  assumption  might  work  to  advantage  in  the 
case  of  arches,  though  similar  conditions  for  a  clear  cold  day  might  prove  more  severe. 

The  painting  of  steel  structures  of  the  indeterminate  class  thus  assumes  considerable 
importance,  since  light  colors  will  obviously  keep  down  the  temperature  while  black 
paint  will  absorb  heat. 

Masonry  structures  are  not  so  severely  distorted  by  temperature  changes,  but  the 
elasticity  of  masonry  being  proportionately  lower,  the  temperature  stresses  may  prove 
equally  dangerous. 

ART.  57.     EFFECT   OF   SHOP   LENGTHS   AND  ABUTMENT  DISPLACEMENTS 

Every  structure  is  designed  for  a  certain  geometric  figure  for  a  condition  of  no 
stress.  This  is  the  figure  which  the  structure  should  present  when  at  a  mean  temperature 
and  when  carrying  no  loads  of  any  kind.  Naturally  the  unavoidable  errors  in  the  shop 
lengths  of  the  members  as  built  and  the  small  inaccuracies  in  the  joints  and  in  the  loca- 
tion of  the  supports  during  erection,  preclude  the  practical  possibility  of  accomplishing 
this  end. 

For  structures  of  the  determinate  class  this  will  have  no  significance,  but  for  indeter- 
minate types,  these  errors  introduce  initial  stresses  which  may  assume  momentous 
proportions. 

To  avoid  such  stresses  requires  the  utmost  care  during  construction  and  erection, 
and  even  then  only  a  partially  satisfactory  outcome  can  be  expected. 

The  difficulties  attending  the  erection  of  a  structure  so  that  no  member  will  be 
stressed  prior  to  the  introduction  of  the  closing  link,  at  the  proper  temperature,  are 
known  to  every  practical  bridge  man. 

In  any  event  the  final  member  should  be  inserted  at  the  calculated  mean  tem- 
perature. If  this  cannot  be  accomplished,  then  the  exact  length  which  this  member 


ART.  57       DESIGN  OF  STATICALLY  INDETERMINATE  STRUCTURES  209 

should  have  must  be  ascertained  by  measuring  the  space  which  is  actually  available 
when  the  structure  is  all  in  place  and  under  no  stress.  Then  applying  proper  tem- 
perature corrections  to  the  entire  structure,  the  required  length  of  the  final  member  must 
be  ascertained  before  attempting  to  force  the  latter  into  its  place. 

When  the  end  supports  are  not  correctly  placed  or  the  entire  structure  has  an  eror 
in  length,  then  all  the  values  S0,  Sa,  etc.,  and  the  redundant  conditions  will  differ  from  their 
computed  values.  If  such  displacements  are  in  the  same  direction  with  a  redundant 
X,  then  the  stress  X  is  affected  thereby,  otherwise  the  principal  system  only  receives 
the  additional  stress. 

Should  a  redundant  member  possess  an  erroneous  length,  then  all  the  members 
will  receive  certain  initial  stresses,  while  errors  in  the  lengths  of  principal  members  would 
change  somewhat  the  geometric  figure,  but  could  not  effect  the  stresses  S0,  Sa,  etc.,  of  the 
principal  system. 

In  dealing  with  abutment  displacements  it  is  necessary  to  distinguish  between  elastic 
changes  which  the-  supports  may  undergo  as  a  result  of  loading,  and  permanent  settlement. 

An  example  of  elastic  displacement  is  given  in  Fig.  45A,  where  the  supporting  column 
CD  will  be  shortened  by  an  amount  db,  which  can  be  determined  and  then  be  included  in 
the  computations  as  for  any  other  member  of  the  structure. 

However,  should  the  pier  at  C  undergo  permanent  settlement,  or  should  one  abut- 
ment, as  B,  be  pushed  out  horizontally  or  settle  vertically,  then  serious  difficulties 
would  arise  unless  these  displacements  could  be  determined  beforehand,  which  is  usually 
quite  impossible. 

In  either  case  the  stresses  occasioned  by  such  displacements  only,  may  be  found 
as  follows,  provided  the  displacements  are  known  or  can  be  estimated  with  reasonable 
accuracy. 

If,  in  Eqs.  (55s),  the  terms  involving  the  temperature  effects  be  dropped,  then  for 
internal  redundancy,  the  displacements  da,  db,  etc.,  are  evaluated  from  Eqs.  (7j),  and  the 
work  of  the  reactions  is  found  as  shown  in  the  derivation  of  Eqs.  (?E). 

When  the  redundancy  is  purely  external  then  the  Eqs.  (55s)  again  apply  by  treat- 
ing the  reactions  R  as  the  reactions  of  the  principal  system  and  evaluating  the  elastic 
displacements  da,  db,  etc.,  and  the  Ar  for  each  reaction  R,  using  such  considerations  as 
given  in  Art.  12,  or  estimating  these  displacements  as  elastic  yielding  in  the  masonry 
supports,  etc.  See  also  examples  in  Articles  49  and  50. 

In  any  case  the  combined  stresses  from  all  causes  are  given  by  Eqs.  (55A)  and  no 
further  comment  is  necessary  here  except  to  emphasize  the  inadvisability  of  adopting 
externally  indeterminate  structures  whenever  immovable  supports  are  not  available.  Even 
for  bed  rock  foundations,  this  will  depend  largely  on  the  depth  to  which  the  masonry 
must  be  carried  and  also  on  the  quality  of  masonry  used. 

When  steel  towers  or  pendulum  piers  are  employed  to  support  an  indeterminate 
structure,  then  the  elastic  deformation  of  such  supports  must  be  considered  in  comput- 
ing the  structural  stresses. 


CHAPTER  XII 
STRESSES  IN   STATICALLY  DETERMINATE  STRUCTURES 

ART.   58.     DEAD    LOAD   STRESSES 

(a)  General  considerations.  The  purpose  of  taking  up  the  question  of  stresses  ir 
statically  determinate  structures  is  not  with  any  intention  of  covering  this  subject  exhaust- 
ively, but  merely  to  present  in  the  briefest  possible  space  the  methods  which  are  besl 
suited  to  the  analysis  of  all  ordinary  structures. 

It  was  not  deemed  advisable  to  take  up  the  present  chapter  before  having  treated 
influence  lines  and  developing  the  general  criteria  for  the  position  of  loads  for  maximum 
and  minimum  stresses  in  Chapter  IV.  The  fundamental  conceptions  there  presented 
are  very  material  to  a  broader  understanding  of  determinate  structures  and  hence  the 
present  order  of  subjects  was  considered  justifiable. 

Since  it  is  desired  to  treat  methods  of  analysis  and  not  types  of  structures,  only 
the  general  case  of  non-parallel  chords  will  be  considered.  Whenever  a  structure  is  sim- 
plified by  introducing  parallel  chords  and  otherwise  simple  relations  between  its  dimen- 
sions, then  naturally  the  analysis  becomes  less  complicated,  until  it  might  be  said  that 
the  problem  is  reduced  to  a  mere  application  of  arithmetic.  The  so-called  algebraic 
methods  are,  therefore,  passed  over  without  further  consideration. 

Since  the  dead  loads  are  fixed  in  position  and  magnitude,  the  stresses  produced  by 
them  in  any  structure  must  be  absolutely  invariable.  The  live  loads,  however,  owing 
to  their  shifting  position,  may  produce  a  variety  of  conditions  and  stresses  which  may 
tend  to  increase  or  to  diminish  dead  load  stresses. 

Therefore,  every  member  may  be  said  to  be  subjected  to  a  maximum  and  a  minimum 
stress  corresponding  to  the  peculiar  or  critical  positions  of  the  live  loads.  These  critical 
positions  are  always  determined  from  certain  tests  or  criteria,  differing  for  different 
members  of  the  same  structure  according  to  the  principles  discussed  in  Articles  20,  23 
and  24. 

It  was  found  there  that  for  the  chord  members  the  maximum  stresses  are 
produced  for  the  case  of  maximum  loading  over  the  entire  span,  while  the  minimum 
stresses  in  these  members  result  from  the  minimum  total  load,  which  is  the  dead 
load. 

This  condition  is  very  different  for  the  web  members,  where  the  dead  load  rarely 
produces  minimum  stresses  because  different  positions  of  partial  live  load  may  pro- 
duce live  load  stresses  of  opposite  signs.  When  these  are  combined  with  the  dead  load 

210 


ART.  58        STRESSES  IX  STATICALLY  DETERMINATE  STRUCTURES 


211 


stresses  to  obtain  total  stresses,  then  the  minimum  total  stresses  are  less  than  the  dead 
load  stresses,  while  the  maximum  total  stresses  exceed  the  dead  load  stresses. 

When  a  web  member  is  designed  to  take  only  one  kind  of  stress  like  tension,  and  this 
stress  is  reversed  by  the  live  loads,  then  the  panel  containing  such  member  must  be 
supplied  with  a  counter  web  member  or  the  original  web  member  must  be  differently 
designed  so  as  to  resist  both  tension  and  compression.  This  latter  is  considered  the 
best  modern  practice. 

Hence,  in  order  that  the  relation  between  the  dead  and  live  load  stresses  may  be 
clearly  brought  out  in  the  analysis,  it  is  always  necessary  to  determine  these  stresses 
separately.  Also,  the  methods  which  apply  to  the  solution  of  dead  load  stresses  are  usually 
not  the  best  adapted  to  finding  the  live  load  stresses,  which  presents  another  reason  for 
dealing  with  the  general  subject  under  separate  headings. 

However,  each  method  given  will  be  applied  directly  to  all  the  members  of  a  struc- 
ture, showing  the  application  both  to  chords  and  web  members  before  proceeding 
further. 

(b)  Aug.  Hitter's  methods  of  moments  (1860).  This  method  is  based  on  the  gen- 
eral theorem  of  moments.  Accordingly  for  all  external  forces  in  the  same  plane  and 
constituting  a  system  in  equilibrium,  the  sum  of  the  static  moments,  about  any  point 
in  this  plane,  must  equal  zero. 

If  now  such  a  set  of  forces  or  loads  be  applied  to  a  frame,  Fig.  58A,  and  these  loads, 
together  with  the  two  end  reactions  A  and  B,  form  a  system  in  equilibrium,  then  if  the 


FIG.  58A. 


frame  be  cut  by  a  section  tt,  the  stresses  in  the  members  cut,  if  acting  as  external  forces, 
must  maintain  equilibrium  of  all  the  external  forces  on  either  side  of  the  section.  If, 
further,  such  planes  can  be  passed  which  do  not  cut  more  than  three  members  of  the 
frame,  then  the  stresses  in  the  members  cut  may  be  determined  one  at  a  time,  by  taking 
as  the  center  of  moments  the  intersection  of  two  of  the  members.  The  moments  of  the 
two  intersecting  members  thus  become  zero. 

The  following  designations  are  used: 

For  member  U  the  center  of  moments  is  n  and  the  ever  arm  is  rn; 

For  member  L  the  center  of  moments  is  m  and  the  lever  arm  is  rm; 

For  member  D  the  center  of  moments  is  i  and  the  lever  arm  is  r£; 

41so  calling  M  the  sum  of  the  moments  of  the  external  forces  A,  P,  and  P2  to  the 


212  KINETIC  THEORY  OF  ENGINEERING   STRUCTURES       CHAP.  XII 

left  of  the  section  it  and  using  subscripts  n,  m  and  i  to  designate  the  centers  about  which 
the  moments  are  taken,  then  for  positive  moments  in  a  clockwise  direction 


Urn+Mn=Q       or       U  =  ~± 

" 


-Lrm+Mm=Q       or 


M. 

-Dri-Mi  =0       or       Z>=-—  -1 


(5SA) 


where  a  positive  stress  indicates  tension  and  a  negative  stress  indicates  compression. 
Thus  a  top  chord  member  is  always  in  compression  and  a  bottom  chord  member  is  always 
in  tension,  while  a  web  member  may  have  either  stress  depending  on  its  inclination  with 
respect  to  its  center  of  moments. 

Hence,  the  stress  S,  in  any  member  of  a  determinate  frame,  may  always  be  expressed 
in  terms  of  a  moment  M  and  its  related  lever  arm  r,  by 


(58B) 


which  is  Ritter's  fundamental  moment  equation  so  extensively  used  in  the  previous 
chapters. 

The  centers  of  moments  for  the  chord  members  are  thus  seen  to  be  located  opposite 
the  chord  in  question,  while  for  a  web  member  the  center  of  moments  may  be  anywhere, 
depending  upon  the  relative  inclinations  of  the  two  chords  composing  the  panel.  The 
lever  arms  should  be  scaled  from  a  large  scale  drawing  or  be  computed. 

When  the  three  members  cut  by  a  section  intersect  in  the  same  point,  the  method 
fails.  Also,  when  the  chords  are  parallel  the  lever  arm  for  the  diagonal  becomes  infinite. 
Hence  the  stress  in  such  a  diagonal  cannot  be  found  by  a  direct  application  of  Eq. 
1.58B)  without  some  modification. 

For  the  case  of  chords  which  are  parallel  or  nearly  so,  the  stress  in  a,  web  member 
may  be  found  by  first  computing  the  stresses  in  the  chords  and  then  finding  the  web 
stresses.  This  is  done  by  choosing  for  the  center  of  moments  for  a  web  member,  any 
convenient  point  of  one  chord  and  then  writing  a  moment  equation,  including  the  other 
chord  of  the  panel  cut  among  the  external  forces. 

However,  there  is  a  simpler  method  of  finding  the  web  stresses  in  structures  with 
parallel  chords  and  that  is  from  the  shear  in  the  panel  cut. 

Thus  in  the  center  panel  mq  the  chords  are  parallel  and  the  diagonal  pq  is  cut  by 
a  section  ft'.  The  lever  arm  for  pq  would  be  infinite. 

Since  the  sum  of  the  moments  of  the  external  forces  to  the  left  of  the  section  and  of 
the  three  members  cut  must  be  zero,  therefore  the  sum  of  the  vertical  components  of 
these  forces  and  stresses  to  the  left  of  the  section  must  likewise  be  equal  to  zero. 

The  sum  of  the  vertical  components  to  the  left  of  the  section  Ft'  is  called  the  shear  Q. 


ART.  58        STRESSES  IN  STATICALLY  DETERMINATE  STRUCTURES  213 

It  is  equal  to  the  end  reaction  A  minus  the  sum  of  the  panel  loads  between  A  and  the  section. 
Thus 


Q=A-*lP,       ..........     (58c) 

where  the  summation  covers  only  the  panel  loads  to  the  left  of  the  section  and  the  shear 
is  taken  positive  upward  on  the  left  of  the  section. 

The  chords,  being  horizontal,  cannot  have  a  vertical  component,  hence  the  only 
stress  in  any  of  the  three  members  cut,  which  can  have  a  vertical  component,  is  the  stress 
in  the  diagonal  pq=D±.  The  vertical  component  of  Dl  is  then  DI  cos  0,  and  the  sum 
of  the  vertical  components  of  all  forces  and  members  cut  on  the  left  of  the  section  W 
becomes 


or 


When  DI  is  vertical,  then  0=0  and  the  stress  becomes  equal  to  the  shear. 

The  above  moment  equation  also  admits  of  a  graphic  solution  which  will  not  be 
considered  here. 

(c)  The  method  of  stress  diagrams.  The  first  description  of  these  diagrams  seems 
to  have  come  from  Bow,  and  in  1864  Clerk-Maxwell  published  a  paper  "  On  Reciprocal 
Figures  and  Diagrams  of  Forces  "  in  which  he  presents  a  scientific  treatment  of  the  sub- 
ject. Cremona,  in  1872,  discussed  the  geometric  properties  of  stress  diagrams,  showing 
their  general  usefulness  in  connection  with  graphostatics. 

English  and  American  writers,  therefore,  call  such  diagrams  "Maxwell  stress  diagrams," 
while  in  Germany  and  France  they  bear  Cremona's  name. 

The  Maxwell  stress  diagram,  so  called  in  the  present  work,  serves  a  most  valuable 
purpose  in  the  graphic  analysis  of  all  determinate  frames,  and  is  generally  applicable 
to  all  cases  which  are  susceptible  to  treatment  by  Ritter's  moment  method. 

A  peculiar  relationship  exists  between  a  frame  and  its  stress  diagram  by  which  each 
member  of  the  frame  is  parallel  to  a  line  of  the  diagram  and  each  pin  point  of  the 
frame  is  represented  by  a  force  polygon  in  the  diagram.  It  is  thus  equally  possible  to 
construct  a  stress  diagram  for  a  given  frame  or  to  construct  a  frame  from  a  given  stress 
diagram.  Hence,  the  term  "  reciprocal  figures  "  used  by  Maxwell. 

The  closed  force  or  funicular  polygon,  which  constitutes  the  basis  for  the  Maxwell 
diagram,  was  known  to  Stevin,  1608,  and  Varignon,  1725,  and  marks  the  beginning  of 
graphics.  Such  a  polygon  may  be  drawn  for  any  set  of  forces  in  equilibrium. 

Since  all  the  forces  meeting  in  a  pin  point  must  constitute  a  system  in  equilibrium,  a 
closed  force  polygon  may  be  drawn  for  each  such  point.  Hence  a  Maxwell  diagram  is 
merely  a  succession  of  closed  force  polygons  drawn  for  all  the  pin  points  of  a  frame. 

For  any  given  frame,  the  directions  of  all  forces  and  members  are  given  or  must  be 
assumed  before  proceeding  to  an  analysis  of  stresses.  Also,  if  the  frame  constitutes  a 
system  in  equilibrium,  then  the  externally  applied  loads  must  be  in  static  equilibrium 
with  the  supports  or  reactions,  and  all  these  in  turn  must  be  in  equilibrium  with  the 
internal  stresses  in  the  members. 


KINETIC  THEORY  OF  ENGINEERING   STRUCTURES         CHAP.  XII 


Any  pin  point  in  equilibrium  and  acted  upon  by  any  number  of  forces  of  known  direc- 
tions, may  then  be  represented  in  the  force  polygon  by  a  succession  of  forces  and  stresses 
respectively  equal  and  parallel  to  the  forces  and  stresses  meeting  in  that  pin  point. 
Since  all  directions  are  known,  two  magnitudes  may  be  found  by  inserting  the  unknown 
members  in  such  a  way  as  to  close  the  force  polygon. 

This  is  the  fundamental  principle  of  the  Maxwell  diagram.  It  is  illustrated  in 
Fig.  58s,  where  the  force  P  and  stresses  S\  and  82  are  known  and  the  stresses  S3  and  S4 
are  of  unknown  magnitudes,  but  of  given  positions.  All  the  forces  meet  in  the  point 
A  and  are  supposed  to  be  in  equilibrium. 

The  method  of  drawing  the  stress  diagram  and  nomenclature  used  in  the  figure  is 
precisely  the  same  in  all  cases  and  affords  an  easy  way  of  deciding  the  direction  of 
action  of  each  unknown  force. 

The  clockwise  arrow  indicates  the  order  in  which  the  forces  are  assembled  in  the 
stress  diagram.  Passing  around  the  point  A  in  this  direction  the  first  known  force  reached 
is  P.  Hence  the  letters  a,  b,  c,  d,  e  are  supplied  as  shown  in  the  angles  between  the 


Forces  Acting  on  A. 


FlG.  58B. 


Stress  Diagram. 


forces,  such  that  the  force  P  is  included  between  a  and  6.  In  speaking  of  the  force  ab 
in  the  stress  diagram,  we  mean  a  force  equal  and  parallel  to  the  force  P  and  acting 
in  the  given  direction  from  a  to  6  in  the  stress  diagram. 

Observing  this  designation,  the  other  known  forces  are  added  in  their  proper  order 
in  the  stress  diagram  and  made  to  act  in  the  given  directions  a-b-c-d  as  found  by  going 
around  the  point  A  in  a  clockwise  direction.  The  stress  diagram  from  a  to  d  is  thus 
obtained  and  may  now  be  closed  by  drawing  a  line  de'  \  S%  through  d  and  another  line 
ae"  ||  84  through  a.  The  intersection  e  thus  found  completes  the  force  polygon  and 
determines  the  stresses  £3  and  S±  both  in  magnitude  and  direction  of  action.  Thus  the 
arrows  around  the  force  polygon  must  be  in  the  same  direction  as  indicated  by  the 
initial  given  force  P.  Accordingly  the  force  83  acts  in  the  direction  from  d  to  e  and  the 
force  S^  acts  in  the  direction  from  e  to  a.  All  forces  in  the  force  polygon  are  laid  off  to 
a  certain  scale  by  which  the  unknown  forces  are  finally  determined.  A  force  or  stress 
acting  away  from  the  pin  point  exerts  a  positive  or  tensile  stress. 

It  is  thus  seen  that  the  unknown  stresses  in  two  of  the  members  meeting  in  any 
pin  point  may  be  determined  by  means  of  a  closed  force  polygon.  However,  if  a  given 
frame  presents  no  pin  point  involving  only  two  unknown  forces,  then  the  method  can- 


ART.  58        STRESSES  IN  STATICALLY  DETERMINATE  STRUCTURES 


215 


not  be  applied  except  by  first  finding  the  forces  in  excess  of  two,  by  some  other  means. 
When  there  are  three  unknown  forces  acting  on  the  point,  and  there  are  no  redundant 
conditions  involved,  then  Ritter's  moment  method  will  always  furnish  the  necessary 
solution  for  one  of  the  unknowns,  prior  to  drawing  the  Maxwell  diagram. 

The  forces  in  Fig.  58B  were  arbitrarily  assembled  in  a  clockwise  direction.  A 
counter  clockwise  direction  might  have  been  chosen  with  equal  right,  though  the  stress 
diagram  would  then  have  occupied  a  symmetric  position  with  respect  to  the  present 

case. 

To   illustrate   the   method,  a  simple  truss,  Fig.    58c,  is  used, 
are  each  10  kips  and  the  reactions  A  =B  =40  kips. 

k 

-43.0 


B-40K. 


The  panel  loads  P 


FIG.  5Sc. 

The  reactions  must  always  be  computed  in   order  to  supply  the  forces  which  are 
necessary  to  establish  a  system  of  external  forces  in  absolute  equihbrium.     The 
diagram  is  then  lettered  and  the  direction  of  assembling  the  forces  is  chosen  so  as  to  locate 
the  stress  diagram  properly  on  the  paper. 

A  pin  point  embracing  only  two  unknown  members  must  be  fleeted  for  a  point 
beginning  and  by  inspection  it  is  seen  that  either  abutmentsaUsfies  this  condition 
theTress  diagram  iscommenced  by  drawingjhe  triangle  «*  inwhich  the  vertica upward 
reaction  «  is  known  and  the  two  forces  cb  and  tajire  found,     lollo^  -ound 
triangle  in  the  direction  a-c-b,  it  is  seen  that  the  force  cb  acts  toward  the 


216 


KINETIC  THEORY  OF  ENGINEERING   STRUCTURES        CHAP.  XII 


The  stress  in  ba  is  found 


produces  negative  or  compressive  stress  in  the  member    cb. 
to  act  away  from  A ,  and  is  thus  positive  or  tensile. 

The  construction  of  the  diagram  is  now  continued  by  assembling  the  forces  around 
pin  point  1,  beginning  with  the  first  known  member  be,  then  adding  the  known  force 
cg  =  Pi  and  finally  closing  the  figure  by  drawing  gf  and/6,  which  are  the  two  unknowns 
for  point  1 . 

The  members  ub  and  bf  being_known_the  closed  force  polygon  for  pin  point  2  may 
next  be  drawn  to  find  the  stresses  fe  and  ed. 


-*TTZ\__ \_LI_U.  _v-    '  - 


FIG.  58o. 

The  process  is  continued,  taking  the  pin  points  in  the  following  order:  A-l-2-3-4-5-6- 
7-8.  As  a  final  check  rp=  B. 

AH  the  stresses  are  then  scaled  from  the  stress  diagram  and  written  on  the  truss 
diagram  with  proper  signs,  —  for  compression  and  +  for  tension. 

The  present  example  shows  a  truss  in  which  the  chord  stresses  are  nearly  all  equal 
and  the  web  stresses  are  small.  It  is  similar  to  the  Pauli  truss  and  is  economical  in  design. 

A  very  interesting  Maxwell  diagram  is  presented  in  Fig.  53e,  where  some  of  the 
external  loading  consists  of  a  moment. 

It  frequently  happens  that  loads  are  not  applied  directly  to  the  pin  points,  in  which 


ART.  58        STRESSES  IN  STATICALLY  DETERMINATE  STRUCTURES 


217 


case  certain  load  concentrations  must  be  effected  before  proceeding  to  draw  a  stress- 
diagram.  The  members  so  loaded  will  usually  receive  a  combination  of  direct  stress  and 
bending  as  illustrated  by  the  member  AC,  Fig.  58D,  representing  a  portion  of  a  roof  truss. 
The  parallel  loads  PI  to  P5,  acting  on  the  member  AC,  may  be  combined  (graphically 
or  analytically)  into  a  resultant  R  of  kncfwn  position  and  magnitude,  and  R  may  then 
be  resolved  into  the  components  R \  and  R%,  constituting  the  pin  point  concentrations 
at  A  and  C  respectively. 


FIG.  58e. 

After  all  loads  have  been  thus  concentrated  on  the  several  pin  points  of  the  struc- 
ture the  total  reactions  may  be  computed  in  the  usual  manner,  taking  into  account  only 
the  loads  R,,  R2,  etc.,  which  now  replace  the  loads  P,  and  then  the  stress  diagram  is 
drawn  in  the  usual  way  and  the  direct  stresses  in  the  members  are  obtained. 

The  reactions  R,  and  R2  are  the  same  as  for  a  span  d,  taken  perpendicular  to  the 
direction  of  the  forces,  and  the  bending  moment  M,  for  any  point  k  distant  x  from  A  , 
may  be  found  from  the  equilibrium  polygon  or  by  computation. 

For  a  direct  stress  -  S,  as  obtained  from  the  stress  diagram,  the  total  thrust  m  the 
member  becomes  N=-S-(Ri  -ft  -ft  -ft)  cos  a.  and  the  moment  M-- 


218  KINETIC  THEORY   OF   ENGINEERING  STRUCTURES        CHAP,  xil 


r]H,  where  H  is  the  pole  distance  measured  to  the  scale  of  forces  and  T?  is 
the  ordinate  of  the  equilibrium  polygon  measured  to  the  scale  of  lengths. 

The  unit  stresses  on  the  extreme  fibers  of  the  column  A  C  then  become  by  Eq.  (49ivi) 


f_, 

J    p±   j  , 


where  y  is  the  distance  of  the  extreme  fiber  from  the  gravity  axis  of  the  section. 

The  dead  weight  of  inclined  or  horizontal  members  would  be  considered  in  this 
manner,  only  that  for  uniform  loads  the  treatment  is  simplified. 

The  example  given  in  Fig.  58E  belongs  to  the  class  previously  pointed  out,  in  which 
not  less  than  three  unknown  members  meet  in  every  pin  point,  and  hence  a  Maxwell 
diagram  cannot  be  constructed  without  first  computing  one  member  by  Ritter's  method 
of  moments. 

The  resultant  R  is  found  as  in  the  previous  figure,  and  supposing  the  support  at  .4 
to  be  a  roller  bearing,  the  reaction  at  R\  must  be  vertical  and  RZ  must  pass  through  B 
and  d.  Hence  the  reactions  of  known  directions  may  be  found  from  the  force  polygon 
by  resolving  R  into  R  \  and  R%. 

The  stress  in  the  member  AB  is  then  found  by  passing  a  section  tt',  cutting  only  three 
members.  Then  with  C  as  center  of  moments  Eq.  (58s)  gives 

AB-H-(Rr-R,lY-(Rr-RrV--R(r-r' 
C"     Kl2)Ji     (™     K2]h     J\r     2 

RI  and  H  being  now  known  the  Maxwell  diagram  can  be  commenced  for  the  pin  point 
A  and  continued  throughout  the  truss. 

For  vertical  loads  the  solution  may  be  conducted  analytically. 

ART.  69.     LIVE   LOAD   STRESSES 

(a)  The  critical  positions  of  a  train  of  moving  loads  to  produce  maximum  and  minimum 
stresses  in  any  member  of  a  given  determinate  structure  must  be  known  prior  to  applying 
any  method  for  finding  the  stresses  themselves.     This  question  was  fully  discussed  in 
Arts.  20,  23  and  24,  and  in  the  General  Considerations  of  Art.  58. 

All  the  methods  for  the  analysis  of  stresses  which  follow  here  presuppose  a  knowl- 
edge of  the  criteria  for  the  positions  of  loads.  The  reader  is  referred  to  the  articles  just 
mentioned  without  repeating  this  discussion  here. 

(b)  The  method  of  influence  lines,  fully  treated  in  Chapters  IV  and  V,  may  be  men- 
tioned as  the  most  universal,  answering  as  it  does  all  questions  relating  to  criteria  for 
positions  of  loads  and  magnitudes  of  stresses  for  any  determinate  structure. 

The  method  requires  no  further  explanation  here  except  to  point  out  the  types  of 
structures  which  warrant  its  application. 

In  general,  the  more  complicated  the  geometric  figure  of  a  structure,  the  greater 
the  advisability  of  employing  the  method  of  influence  lines,  since  the  geometric  relations 
of  the  truss  dimensions  are  thereby  expressed^,  independently  of  the  loads. 


ART.  59        STRESSES  IN  STATICALLY  DETERMINATE  STRUCTURES  219 

Hence  it  would  scarcely  be  advisable  to  apply  influence  lines  to  any  truss  with  parallel 
chords  except  in  cases  of  complicated  cantilever  systems.  On  the  other  hand  a  saving 
of  labor  might  be  expected  in  analyzing  structures  in  which  either  or  both  chords  are 
curved.  The  method  certainly  offers  obvious  advantages  in  all  cases  illustrated  in 
Chapter  V,  especially  when  concentrated  loads  are  employed. 

It  may  be  added  that  any  structure  which  does  not  warrant  the  accurate  analysis 
by  concentrated  load  systems  should  not  be  dignified  to  the  extent  of  being  called  a 
modern  bridge.  Nor  should  anyone  not  familiar  with  these  more  exact  methods  be 
intrusted  with  the  design  of  bridges. 

It  should  also  be  emphasized  that  all  computations  of  stresses  should  err  decidedly 
on  the  side  of  safety,  since  the  secondary  stresses  produced  by  the  friction  on  pin-con- 
nected joints  are  frequently  a  very  considerable  quantity,  to  say  nothing  of  the  failure 
to  analyze  properly  the  effect  of  the  riveted  connections. 

(c)  Discussion  of  methods  in  common  use.     In  general,   all    methods    involve    the 
solution  of  the  following  distinct  problems:    To  find  certain  moments,  shears  and  end 
reactions  resulting  from  certain  critical  positions  of  the  train  of  loads  and  then  to  deter- 
mine the  maximum  and  mimimum  stresses  in  a  particular  member  in  terms  of  these  moments 
or  shears. 

Both  analytic  and  graphic  solutions  have  been  proposed  to  solve  all  of  these  prob- 
lems in  their  various  phases.  Some  are  applicable  only  to  trusses  with  parallel  chords  while 
others  are  more  general.  Still  others  are  used  when  the  position  of  the  train  is  chosen 
so  as  to  avoid  the  complications  arising  from  the  exact  loadings  required  by  the  criteria 
for  maximum  and  minimum  stresses  in  a  certain  member. 

It  would  be  outside  the  province  of  the  present  chapter  to  present  all  of  these 
methods  in  detail,  though  many  of  them  are  exceedingly  interesting  and  ingenious.  The 
author  would  offer  the  general  criticism  that  the  methods  in  common  use  are  too  special 
in  their  application,  and  when  a  more  general  problem  is  encountered  no  one  method 
will  suffice,  but  several  methods  must  be  combined,  thus  necessitating  an  intimate  knowl- 
edge of  many  limited  methods  to  accomplish  a  complete  analysis. 

The  following  method  is  an  attempt  to  determine  the  live  load  stresses  in  each  of 
the  members  of  any  simple  truss  by  applying  the  same  process  throughout. 

(d)  The  author's  method  of  determining  live  load  stresses  is  based  on  Ritter's  com- 
prehensive and  universal  method  of  moments.     The  moment  of  the  live  loads  on  one 
side  of  the  section  taken  about  the  center  of  moments  for  any  particular  member  is 
derived  from  the  principles  of  the  sum  A  line,  see  Fig.  22fi. 

Thus  the  stress  in  any  member  of  any  determinate  truss  is  given  by  Eq.  (58s)  as 


except  for  the  web  members  of  a  truss  with  parallel  chords  where  the  stress  is  derived 
from  the  shear  according  to  Eq.  (58o)  as 


(59B) 


220 


KINETIC   THEORY   OF   ENGINEERING  STRUCTURES        CHAP.  XII 


The  values  of  M  and  Q  for  any  position  of  a  live  load  will  now  be  derived  and 
evaluated  first  from  the  sum  A  line,  and  then  from  a  single  tabulation  of  moments. 

The  critical  position  of  the  loads  for  any  particular  effect  on  a  certain  member  is 
supposed  to  have  been  determined  in  accordance  with  Arts.  20,  23  and  24. 


M   1 P  il  11 1 


1 

*B 

,  «•  —  *m  —  - 

1               lam                    -        ., 

...  |    -                               _, 

FIG.  59A. 


Referring  to  Fig.  59A  and  calling  MB  the  sum  of  the  moments  of  all  loads  on  the  span 
about  the  reaction  B  and  Mm  the  sum  of  the  moments  about  any  point  m;  also  calling 
Qm  the  vertical  shear  at  the  point  m;  then  for  loads  1  to  q}  covering  a  distance  xq=ljr(l 

!*P;  •  (59c) 


(59D) 

(59E) 
(59r) 


or 


Mm=lam(A- 


lam 


(59c) 


wherein  Aam  =  £3.  Pe/lam  may  be  defined  as  the  end  reaction  at  A  of  the  loads  1  to  m 
covering  the  distance  xm  of  a  simple  span  Zam. 

The  values  A  and  ^4aw  may  be  obtained  from  a  reaction  summation  influence  line, 
or  sum  A  line,  drawn  for  the  reaction  A  and  span  I,  according  to  Art.  22,  Figs.  22A  and 
22s.  The  ordinates  of  the  sum  A  line  may  be  computed  from  Eq.  (59o),  or  constructed 
graphically  by  the  method  given  in  Art.  22. 

Such  a  sum  A  line  is  shown  in  Fig.  59s,  drawn  for  a  span  of  200  feet  and  using  a  train 
of  Cooper's  EGO  loading,  consisting  of  two  locomotives  followed  by  a  uniform  load  of  3000 
Ibs.  per  linear  foot  per  rail. 

The  ordinate  yx  under  the  first  load  PI  at  x,  represents  the  end  reaction  A  for 
the  position  of  loads  indicated,  when  the  span  A  B  is  loaded  for  a  distance  61  from  B. 

The  reaction  Aam,  for  the  same  train  covering  the  distance  xm  from  m  on  the 
span  lam>  may  be  obtained  from,  the  same  sum  A  line  as  follows:  Had  the  sum  A  line 


ART.  59        STRESSES  IN  STATICALLY  DETERMINATE  STRUCTURES 


221 


been  drawn  for  a  span  lam  then  the  ordinate  at  xm  from  B'  would  be  the  required  reaction 
Aam.  However,  this  polygon  was  actually  drawn  for  a  span  I,  and  hence  the  ordinate 
i)xm  distant  xm  from  B'  is  l/lam  times  too  small,  therefore, 


L 


(59n) 


This  illustrates  how  a  sum  A  line  drawn  for  any  span  may  be  used  to  obtain  end 
reactions  for  any  other  span.     However,  in  order  to  avoid  the  multiplication  of  all  ordi- 


o 

«  . 


SCALE  OF  LOADS. 


300  KIPS. 


222  KINETIC   THEORY  OF  ENGINEERING  STRUCTURES        CHAP.  XII 

nates  by  I,  it  is  best  to  draw  the  sum  A  line  for  a  span  of  1000  ft.,  and  then  derive  all 
ordinates  according  to  Eq.  (o9n)?  by  which  the  ordinates  of  one  A  line  are  to  those  of 
another  A  line  in  the  inverse  proportion  of  the  spans  for  which  the  lines  were  drawn. 

A  portion  of  such  a  1000  ft.  sum  A  line  is  shown  in  Fig.  59B,  from  which  the  actual 
reactions  A  and  Aam,  for  a  span  /=200  ft.,  are  obtained  as 


and       ^=    .......     (59.) 


I 

where  JJA  is  the  ordinate  distant  61  from  B'  and  f)xm  is  the  ordinate  distant  xm  from  B'. 

Any  moment  MTO  or  chord  stress  is  now  readily  found  from  Eq.  (59o) ,  using  the  values 
from  Eq.  (59i)  as  follows: 

la 


,    ......     (59j) 

and  the  stress  in  any  chord  member  having  the  point  m  for  its  center  of  moments  with 
a  lever  arm  rm,  becomes  by  Eqs.  (59A)  and  (59j) 

(59K) 


The  web  stresses  for  trusses  with  non-parallel  chords  are  derived  from  moments  Mi 
of  the  external  forces  about  the  center  of  moments  i  for  any  particular  web  member. 

Calling  the  lever  arm  rt-  for  a  web  member  with  moment  center  i  distant  Zat-  to  the 
left  of  A  then  for  xm<d  and  Anm  =  ' 


Mi  =  Alai     A  nm  (Lai  +  lan)  , 
or  M^lOOO'-^l^^+U,     ....    V  '..   .  •-  .  (59D 


when  xm>d,  a  case  which  happens  very  rarely,  indicating  that  some  loads  extend  to  the 
left  of  n,  then 


•*"» — -"-w     -Anmv/ai  -rlan)        ^  ^    r(lni 

**^  x 


wherein  the  two  last  terms  must  be  computed.  It  should  be  remembered  that  this  last 
case  does  not  occur  more  than  once  or  twice,  if  at  all,  for  a  whole  analysis  and  hence 
involves  only  a  slight  amount  of  work. 

In  any  case  Eq.  (59A)  gives  the  stress  in  a  web  member  for  non-parallel  chords,  as 


(59x) 


The  web  stresses  for  trusses  with  parallel  chords  are  derived  from  the  shear  accord- 
ing to  Eq.  (59E) 


ART.  59        STRESSES  IN  STATICALLY  DETERMINATE  STRUCTURES  223 

Thus  the  vertical  shear  to  the  left  of  a  section  through  any  panel  ™    when  there 
are  no  loads  to  the  left  of  n,  making  xm<d,  becomes 

0-1     A          1000>M      1000^m 

ym^A-Anm=  — ^         ~^~, (59o) 

wherein  Anm  is  the  reaction  at  n  for  the  loads  in  the  panel  ^m  extending  over  a  distance 
xm  on  a  span  d. 

When  xm>d,  which  would  be  the  rare  case  for  one  or  two  loads  to  the  left  of  n,  then 


r  -2,R-3~2,,P>    •  •  <w 

where  the  two  last  terms  would  have  to  be  computed  for  the  loads  between  x  and  m  as 
in  the  previous  case. 

The  stress  in  any  web  member  of  a  truss  with  parallel  chords  is  then  found  from 
Eq.  (58o)  as 

S=Qmsecd, .     .     (59Q) 

where  Qm  is  given  from  Eqs.  (59o)  or  (59p). 

The  above  demonstration,  necessitating  the  use  of  one  sum  A  line  drawn  for  a  span 
of  1000  ft.,  was  used  only  for  the  development  of  the  formula?.  In  practice,  the  sum 
A  line  is  dispensed  with  by  tabulating  the  ordinates  1000rM  for  any  case  of  standard 
loading. 

One  such  table  (see  Table  59A) ,  will  then  suffice  for  the  computation  of  all  chord  and 
web  stresses  in  any  simple  determinate  truss  for  the  general  case  of  moving  train  loads. 

Table  59A  thus  represents  the  values  of  1000A  =1000^  =MB=the  sum  of  the 
moments  of  all  loads  on  the  1000  ft.  span  about  the  reaction  B,  expressed  in  kip  feet. 
These  moments  are  readily  computed  from  Eqs.  (59c),  and  (59i),  thus 

4/5  =  1000^4=--  VVp  +  /3VV,  . (59R) 


where  lq  is  the  distance  the  train  is  moved  ahead  whenever  a  new  load  P4  is  brought 
on  the  span. 

Eq.  (59n) ,  holds  until  the  uniform  load  reaches  the  point  B,  after  which  the  moments 
must  be  figured  as  for  uniformly  distributed  loads.  The  table  suggests  the  ease  of 
computation  and  might  be  extended  over  the  whole  span  of  1000  feet,  though  for  pur- 
poses of  illustration  this  was  not  considered  necessary. 

To  facilitate  interpolation,  the  differences  in  MB  for  one  foot  were  added  in  a  separate 
column.  Interpolations  for  distances  between  those  given  are  thus  readily  performed 
and  these  are  strictly  accurate  up  to  the  point  where  the  uniform  load  is  reached.  After 
that  there  is  a  slight  error  because  the  moment  for  uniformly  distributed  loads  varies 
as  I2.  However,  the  error  from  this  source  is  entirely  negligible.  

Example.  Required  the  maximum  live  load  stresses  in  the  chord  U2U3,  and  in  the 
diagonal  ~U^LZ  of  the  200  ft.  truss  shown  in  Fig.  59e,  using  Cooper's  E  60  loading. 


224 


KINETIC  THEORY  OF  ENGINEERING  STRUCTURES        CHAP.  XII 


TABLE  59A. 
REACTIONS   FOR   COOPER'S  E-60   LOADING,   SPAN   OF   1000  FEET. 


Wheel 
No. 

Load 
Lenj^th, 

6, 

Feet. 

Total  Load 

S?P 

Kips. 

MB=IQOQT)A 

Kips. 

Diff.  for 
1  Foot. 

Kips. 

Wheel 
No. 

Load 
Length, 

&1 
Feet. 

Total  Load 

S?P 

Kips. 

-1/5=1000^ 
Kips. 

Diff.  for 
1  Foot, 

Kips. 

523.5 

1 

0 

15.0 

0.0 

26 

144 

531 

41,293.5 

15.0 

538.5 

2 

8 

45.0 

120.0 

27 

149 

546 

43,986.0 

45.0 

553.5 

3 

13 

75.0 

345.0 

28 

154 

561 

46,753.5 

75.0 

568.5 

4 

18 

105.0 

720.0 

29 

159 

576 

49,596.0 

105.0 

583.5 

5 

23 

135.0 

1,245.0 

30 

164 

591 

52,513.5 

135.0 

598.5 

6 

32 

154.5 

2,460.0 

31 

169 

606 

55,506  .  0 

154.5 

613.5 

7 

37 

174.0 

3,232.5 

32 

174 

621 

58,573.5 

174.0 

628.5 

8 

43 

193.5 

4,276.5 

33 

179 

636 

61,716.0 

193.5 

643.5 

9 

48 

213.0 

5,244.0 

34 

184 

651 

64,933.5 

213.0 

658.5 

10 

56 

228.0 

6,948.0 

35 

189 

666 

68,226.0 

228.0 

673.5 

11 

64 

258.0 

8,772.0 

36 

194 

681 

71,593.5 

258.0 

688.5 

12 

69 

288.0 

10,062.0 

37 

199 

696 

75,036.0 

288.0 

703.5 

13 

74 

318.0 

11,502.0 

38 

204 

711 

78,543.5 

318.0 

718.5 

14 

79 

348.0 

13,092.0 

39 

209 

726 

82,146.0 

348.0 

733.5 

15 

88 

367.5 

16,224.0 

40 

214 

741 

85,813.5 

367.5 

748.5 

16 

93 

387.0 

18,061.5 

41 

219 

756 

89,556.0 

387.0 

763.5 

17 

99 

406.5 

20,383  .  5 

42 

224 

771 

93,373.5 

406.5 

778.5 

18 

104 

426.0 

22,416.0 

43 

229 

786 

97,266.0 

426.0 

793.5 

19 

109 

426.0 

24,546.0 

44 

234 

801 

101,233.5 

433.5 

808.5 

20 

114 

441.0 

26,713.5 

45 

239 

816 

105,276.0 

448.5 

823.5 

21 

119 

456.0 

28,956.0 

46 

244 

831 

109,393.5 

463.5 

838.5 

22 

124 

471.0 

31,273.5 

47 

249 

846 

113,586.0 

478.5 

853.5 

23 

129 

486.0 

33,666.0 

48 

254 

861 

117,853.5 

493.5 

868.5 

24 

134 

501.0 

36,133.5 

49 

259 

876 

122,196.0 

508.5 

883.5 

25 

139 

516.0 

38,676.0 

50 

264 

891 

126,613.5 

All  loads  in  kips  for  one  rail. 


ART.  59        STRESSES  IN  STATICALLY  DETERMINATE  STRUCTURES  225 

For  the  chord  U2U3,  the  center  of  moments  is  at  m  and  the  maximum  stress  occurs 
when  the  span  is  fully  loaded,  making  6t  =1  =  200  ft.,  and  xm  =lam  =60  ft.    Also  rm  =34.5  ft. 

Then  Table  59A  gives  for  200  ft.,  1000yjA  =75036  +703.5  =75739.5,  and  for  xm  =60  ft 
1000^™  =6948  +4X228  =7860,  whence  by  Eq.  (59K), 

=  J-  [75739.5(1^  -7S60J  =430.78  kips. 


For  the  diagonal  U2L3  with  load  divide  at  zw  =  10.5  ft.,  making  6t  =  140  +  10.5 
=  150.5  ft.,  Zan=40  ft.,  Zat-=98  ft.,  d=20  ft.,  and  r^  =  132  ft.,  and  thus  xm<d,  Table  59A 
gives  1000r?A  =43986  +  1.5X553.5  =44816.3  and  1000^  =  120+2.5X45=232.5. 

Hence  by  Eqs.  (59L)  and  (59N). 


The  minimum  stress  in  C/2^3  may  be  found  by  treating  the  symmetric  member  L7U8 
in  the  right  half  of  the  span. 

Similarly  the  stress  in  any  member  of  the  truss  may  be  determined  for  any  desired 
position  of  the  train  of  loads,  all  by  means  of  the  one  Table  59A. 

It  is  believed  that  the  ease  and  simplicity  of  applying  the  above  method,  together 
with  its  universal  applicability,  should  commend  itself  to  the  practical  designer. 

Certainly  it  does  not  seem  warranted  to  employ  approximate  methods  when  an 
exact  solution  is  possible  without  additional  labor. 


CHAPTER  XIII 
SECONDARY  STRESSES 

ART.   60.     THE    NATURE    OF   SECONDARY   STRESSES 

In  the  previous  chapters  a  framed  structure  was  regarded  as  a  system  of  individual 
members  linked  together  by  frictionless  pin  connections.  It  was  also  supposed  that  the 
neutral  or  gravity  axes  of  all  members  meeting  in  a  panel  point  actually  intersected  in 
one  point.  The  stresses  computed  on  this  basis  are  called  primary  stresses. 

Owing  to  the  friction,  which  always  exists  in  pin-connected  joints,  and  the  rigidity 
introduced  by  riveted  connections,  this  condition  is  never  fully  realized  in  practice.  Hence 
certain  moments  or  bending  effects  are  always  produced  in  the  proximity  of  the  connected 
ends,  which  set  up  bending  stresses  of  variable  character  and  magnitude.  These  are 
generally  called  secondary  stresses  as  distinguished  from  the  primary  stresses. 

All  structures  involving  redundancy  are  subjected  to  elastic  deformations  caused 
by  the  loads,  temperature,  and  abutment  displacements  giving  rise  to  what  may  be  called 
additional  stresses.  These  also  exist  to  a  slight  degree  in  determinate  structures,  especially 
when  their  geometric  shape  is  peculiarly  subject  to  large  load  and  temperature  deflec- 
tions as  in  the  case  of  three-hinged  arches.  Still  other  causes,  such  as  erroneous  shop 
lengths  of  members,  wind  pressure,  and  impact  and  brake  effects  of  moving  loads,  pro- 
duce stresses  belonging  to  this  class  which  do  not  admit  of  exact  determination  and  can 
at  best  be  merely  estimated. 

The  first  theoretical  discussions  of  secondary  stresses  were  given  by  Asimont,  1877; 
Manderla  and  Engesser,  1879,  and  Winkler,  1881.  Graphic  solutions  were  published  by 
Landsberg,  1885;  W.  Ritter,  1890;  and  Mohr,  1891-3.  A  resume  of  most  of  these  is  given 
by  C.  R.  Grimm,  C.  E.,  "  Secondary  Stresses  in  Bridge  Trusses,"  1908. 

The  facts  which  were  revealed  by  these  discussions  gave  rise  to  very  material 
improvements  in  the  modern  practices  of  designing  structures.  Thus,  the  details  of  con1 
nections  at  panel  points  and  between  floorbeams  and  truss  members  have  been  vastly 
improved,  and  particular  attention  is  now  being  given  to  a  more  judicious  design  of  the 
members  themselves,  especially  those  subjected  to  compressive  stress. 

It  is  not  the  purpose  here  to  treat  all  the  various  methods  of  calculating  secondary 
stresses,  but  merely  to  present  what  appears  to  the  author  as  the  most  understandable 
and  usable  method.  The  nature  of  the  problem  is  such  as  to  preclude  the  possibility 
of  considering  all  the  complexities  involved  and  yet  remain  within  the  limits  of  practical 
utility.  Hence  some  approximations  must  be  applied,  and  justly  so,  because  many  of 
the  influences  are  too  small  to  warrant  the  labor  involved  in  their  analysis,  and  others 
are  beyond  reasonable  limits  of  estimation. 

226 


ART.  Gl  SECONDARY  STRESSES  227 

The  three  primary  causes  for  secondary  stresses  in  bridge  members  are:  the  weights 
of  the  members  themselves  producing  deflections  in  all  except  vertical  members  and  giving 
rise  to  bending  stresses;  the  absence  of  frictionless  panel  point  connections;  eccentric 
connections  between  members  meeting  in  a  common  panel  point. 

Riveted  connections  are  here  treated  like  fixed  ends  so  that  the  elastic  deformations 
of  the  structure  are  supposedly  taken  up  by  flexure  in  the  members  themselves  with- 
out producing  any  changes  in  the  angles  at  the  panel  points.  While  this  assumption  is 
not  absolutely  true,  because  the  bending  moments  cannot  be  resisted  without  producing 
some  elastic  angular  distortions,  yet  it  is  on  the  side  of  safety  and  tends  to  compensate 
some  other  factors  which  are  entirely  neglected. 

Pin-connected  members,  according  to  more  recent  experience,  cannot  be  regarded 
as  free  from  bending  moments  on  account  of  the  rather  excessive  friction  which  always 
prevails  on  the  pins.  In  the  case  of  eye-bars  this  may  add  considerable  bending  stress 
owing  to  the  slenderness  of  such  members.  This  should  not  be  construed  to  mean 
that  pin  connections  are  inferior  to  riveted  joints,  but  that  the  former  do  not  possess 
all  the  advantages  usually  attributed  to  them.  The  distinct  superiority  of  pins  is  to  be 
found  in  the  prevention  of  eccentric  transmission  of  the  direct  stresses,  a  condition  which 
is  difficult  to  accomplish  in  riveted  connections. 

The  excessive  bending  stresses  carried  from  floor-beam  connections  to  the  vertical 
trusses  can  hardly.be  estimated  and  should  be  avoided  as  far  as  possible  by  the  introduc- 
tion of  flexible  connections  in  preference  to  the  rigid  type  so  much  used  in  the  past. 

The  secondary  stresses,  which  will  now  be  considered,  are  those  caused  by  the  elastic 
deformations  taking  place  in  the  plane  of  the  frame  itself  and  due  to  the  three  primary 
causes  above  enumerated. 


ART.  61.     SECONDARY   STRESSES   IN   THE   PLANE    OF   A   TRUSS   DUE   TO 

RIVETED   CONNECTIONS 

Every  elastic  structure  must  undergo  certain  deformation  when  subjected  to  loads. 
If  the  structural  members  are  rivet-connected  at  the  ends,  then  this  deformation  is 
taken  up  by  the  members  themselves  in  the  production  of  certain  bending  moments  which 
are  resisted  by  the  rigid  panel  connections. 

Figs.  6lA  and  61  B  show  the  several  possible  combinations  of  single  and  double  flexure 
with  compression  and  tension  as  they  might  occur  in  any  structure. 

The  straight  chord  AB,  in  each  case,  represents  the  position  of  the  member  on  the 
supposition  of  frictionless  pin  connections,  while  the  curved  line  indicates  the  member 
as  it  would  be  distorted  by  the  bending  moments  and  axial  forces  resulting  from  the  fixed 
riveted  connections. 

Assuming  such  a  member  to  be  released  at  the  ends  by  sections  passed  close  to  the 
panel  points,  then  the  internal  stresses  may  be  replaced  by  external  forces  Pa  and  Pb. 
Each  of  these  may  be  resolved  into  two  forces  and  a  moment  acting  at  the  original  fixed 
ends.  Thus  Pa  is  equivalent  to  Sa,  A  and  a  moment  Ma  while  Pb  is  replaced  by  Sb,  11 
and  a  moment  Mb.  The  conditions  of  equilibrium  require  that  Pa=Pb  and  hence  that, 


228 


KINETIC  THEORY  OF  ENGINEERING  STRUCTURES       CHAP.  XIII 


Sa=Pacosd=Sb=Pbcos6;     also   A=B=P   sin    6    and    Ma+Mb-Al=0.     Since    0    is 
always  very  small  Sa—Sb=Pa-=Pb,  which  assumption  is  entirely  permissible. 

In  the  case  of  compression  members  the  concave  side  of  the  elastic  curve  is  always 


FIG.  6lA — Tension  Members. 

toward  the  resultant  P,  and  hence  a  maximum  moment  may  occur  at  some  point  inter- 
mediate between  A  and  B.  For  tension  members  the  convex  side  is  toward  the  force, 
and  the  maximum  moment  must  occur  at  one  end  of  the  member. 


FIG.  61s — Compression  Members. 


In  any  case  then,  the  stresses  produced  in  a  member  by  the  bending  moments  M« 
and  Mb,  are  readily  determined  provided  these  moments  are  known.     Of  the  several 


ART.  61 


SECONDARY  STRESSES 


229 


methods  proposed  for  the  evaluation  of  the  end  moments,  the  analytic  method  of  Professor 
Fr.  Engesser  is  probably  the  most  direct  and  is  the  one  given  here. 

For  any  structure  of  w  members  there  will  be  2m  unknown  end  moments  to  be 
evaluated.  These  are  derived  from  the  angular  deflections  r,  which  the  respective  fixed 
ends  of  the  members  undergo  as  a  result  of  certain  distortions  Ja,  J/?,  Jf,  etc.,  produced 
in  the  angles  of  the  frame  by  the  work  of  the  external  forces  and  reactions,  as  per  Eqs.  (37n). 

The  first  step  in  the  development  of  the  method  is  then  to  derive  the  relation  between 
the  moments  Ma  and  MI,  and  the  deflection  angles  ra  and  rb  for  a  column  on  two  fixed 
supports. 

The  following  assumptions  are  made  in  the  interest  of  simplifying  the  theory  for 
practical  applications. 

Since  in  well  designed  members  the  shear,  direct  thrust  and  temperature  have  a  very 
slight  effect  on  flexure,  these  factors  are  neglected.  However,  for  poorly  designed  members 
with  small  moments  of  inertia  and  slender  dimensions,  the  deflections  due  to  normal 
thrust  etc.,  may  become  very  considerable  and  these  assumptions  might  not  hold. 

It  is  further  assumed  that  the  axes  of  all  the  members  are  situated  in  the  same 
plane,  and  that  these  axes  are  centric  for  all  members  meeting  in  a  point.  Also,  that  the 
external  loads  are  all  applied  at  the  panel  points,  neglecting  for  the  present  the  flexure 
produced  in  the  members  by  their  own  weight. 

Since  the  fixed  ends,  Fig.  61c,  do  not  of  themselves  produce  any  direct  thrust  or  stress 
in  the  member  ~AB  (not  being  absolutely  fixed  in  space),  therefore,  the  column  involves 
only  two  redundant  conditions  which  are  the  moments  Ma  and  Mb.  The  principal  system 
is  then  a  beam  on  two  supports. 


FIG.  61c. 

The  effect  of  the  direct  stress  S  is  not  considered  here  because  for  well  designed 
members  with  comparatively  large  moments  of  inertia  this  was  found  by  Manderla  to 
be  entirely  negligible.  Otherwise  a  moment  Sij  would  enter  the  general  moment  as 
given  by  Eq.  (6lA). 

Eq.  (7 A)  now  gives  for  A0=0,  An  =  l/l  and  Ab=  -l/l, 

A=A0-AaXa-AbXb=-^+^, 


and  the  moment  Mm,  about  any  point  ra,  becomes 


Mm  = 


a=    (Xb  -  Xa)  +  Xa, 


(61A) 


230 


KINETIC  THEORY  OF  ENGINEERING   STRUCTURES     CHAP.  XIII 


giving 


™_       x  ,      'dMm_x 

-  ~  I  -dXb   ~T 


Then  according  to  Eq.  (15L),  the  virtual  work  of  the  moment  Ma  is 


~j7rr  I  "nXl  ~-^Xa  +  lXa  —  -^Xb  +  -^Xa  —  —Xa  —  £-rjj(2Xa  +  Xb)  . 

EiL\_Zi  Zi  o  o  ^  'QtiiL 

A  similar  process  will  furnish  rb  and  calling  Xa=Ma  and  Xb=Mb,  then  for  any 
member  of  single  curvature 


and 


(61n) 


When  the  moments  Ma  and  Mb  act  in  the  same  direction  then  the  elastic  curve 
will  have  a  point  of  counterflexure  and  a  similar  derivation  to  the  one  just  given  will 
furnish  the  values 

and       Tb=-^=-ft2Mb  —  Ma) (61c) 

6EI 


Considering  a  triangle  as  the  fundamental  element  of  a  frame,  the  distortions  Act, 
Aft  and  Af  in  thethree  angles  resulting  from  the  changes  in  the  lengths  of  the  members, 
expressed  in  terms  of  unit  stresses,  were  determined  in  Art.  37.  It  is  now  necessary  to 
show  the  relation  between  these  angle  distortions  and  the-  deflection  angles  T  at  each  end 
of  each  member  forming  the  triangle. 


FIG.  61o. 


Referring  to  Figs.  6lD  and  6lE,  it  is  easily  seen  that 


(6lD) 


Fig.  6 ID  represents  an  impossible  condition  because  at  least  one  member  of  the 
three  composing  the  triangle  must  undergo  double  curvature  as  shown  by  Fig.  61  E  and 
as  required  by  Eq.  (3?E) ,  whereby 

0.     . (61s) 


ART.  61 


SECONDARY  STRESSES 


23  1 


The  elastic  distortions  Aa,  J/3  and  Jj-  are  given  by  Eqs.  (37o),  and  may  be  considered 
known  for  all  angles  of  a  frame  and  for  one  simultaneous  case  of  loading  and  stress. 

While  Eqs.  (37D)  were  derived  on  the  assumption  of  frictionless  pin  connections 
between  all  members,  they  are  now  used  to  determine  the  amount  of  distortion  in  each 
angle  which  is  prevented  from  taking  place  by  virtue  of  the  rigid  riveted  connections. 

Two  equations  of  the  form  of  Eqs.  (61c),  may  be  written  for  each  side  of  a  triangle 
or  for  each  member  of  a  frame.  Using  the  designations  indicated  by  the  double  sub- 
scripts in  Fig.  6lD,  where  the  first  refers  to  the  apex  or  panel  point  and  the  second  to 
the  far  end  of  the  adjacent  member,  then  by  inserting  the  deflection  angles  r  into  Eqs. 
(6  ID),  the  following  formulae  are  obtained  for  any  triangle  ABC'. 


QEJa  = 


.     .  (6lF) 


cb  +rca)  =        (2MA  +Mbc)  + 

*a  lb 


+Mac) 


In  the  Eqs.  (61r),  the  six  moments  are  unknown,  and  hence  a  solution  is  impossible 
unless  at  least  three  of  them  may  be  independently  derived  from  other  conditions  to  be 
developed  later. 

According  to  Eqs.  (Glo)  the  sum  of  the  three  angle  distortions  J  for  any  triangle, 
must  equal  the  sum  of  the  six  deflection  angles  r  and  by  the  condition  of  Eq.  (6lE),  both 
sums  must  then  be  equal  to  zero,  hence 


(61c) 


Also,  for  any  apex  angles  meeting  in  the  same  panel  point,  the  sum  of  the  angle 
distortions  J  must  equal  the  sum  of  the  deflection  angles  r,  hence  for  any  panel  point 

2J  =  2r,  ........     ....     (6lH) 


and  therefore,  if  one  of  the  deflection  angles  r  is  known,  the  others  around  that  panel 
point  may  be  successively  found  by  applying  equations  of  the  form  of  Eqs.  (6lD). 

For  any  determinate  frame,  there  will  thus  be  as  many  unknown  deflection  angles 
-  as  there  are  panel  points,  and  for  p  panel  points  there  will  be  p  equations  of  the  form 
of  Eqs.  (61n),  each  involving  one  unknown  deflection  angle  T. 

The  following  values 


Mablc 


r  J 

6/c 


•"*•  ca'b 


(61 1) 


232  KINETIC  THEORY  OF   ENGINEERING  STRUCTURES       CHAP.  XIII 

are  now  substituted  into  Eqs.  (61r),  as  a  matter  of  convenience,  to  obtain  the  following 
formulae  for  any  triangle  ABC,  thus 

EAa=Kba+2(Kac+Kab)+Kea,  .  .  .  /  ,~:  .  .  (61j) 
EAp=Kab+2(Kbc+Kba)+Kcb,  .  .  ,  .'._.  .  (61  K) 
EAr=Kbc+2(Kcb+Kca)+Kac.  .  t  -.  \  .  [.  .  "..  (61L) 

Adding  these  three  equations  and  imposing   the  condition  of  Eq.  (6  IE),  then  for 
any  triangle  ABC, 

=Q.      ....,'-.,     (6lM) 


Assuming  now  that  the  three  quantities  K^,  K^  and  Kba  are  known,  then  the  remain- 
ing three  may  be  obtained  from  the  above  equations  as  follows: 

from  Eq.  (61J),  Kca=EJa-2(Kac+Kab)  -K^,     .     .>.     .     ..    (Glx) 

from  Eqs.  (6  IK)  and  (<ol^)  ,  K^^E^+K^-K^+K^,      .     .     .     ;     .     .'    (61o) 
from  Eqs.  (6lL)  and  (6lM),  Kcb=EAr+kab+Kba-Kca.     .     t     .     .     .     .     (61r) 

By  the  conditions  of  static  equilibrium,  the  sum  of  all  the  moments  about  any  panel 
point  must  be  zero,  from  which  p  equations  of  the  following  form  are  obtained,  using 
the  values  from  Eqs.  (61i).  Thus,  for  any  panel  point  A 

~   '•' 


including  all  the  members  meeting  in  this  point.  See  Fig.  61n  and  Eq.  (61u)  for  the 
manner  of  forming  these  equations. 

The  following  procedure  will  furnish  a  solution  for  all  the  bending  moments  of  a  frame, 
using  Eqs.  (6 IN)  to  (61q) :  In  the  first  triangle,  evaluate  three  of  the  quantities  K  in 
terms  of  the  other  three,  using  Eqs.  (61  N,  o,  P).  Two  of  the  values  K  so  found,  also 
belong  to  the  adjacent  or  second  triangle  and  a  third  value  K  in  this  second  triangle 
may  be  found  from  Eq.  (6lQ),  making  three  values  known  to  find  the  other  three  by 
applying  Eqs.  (6lN,  o,  P)  to  the  second  triangle. 

This  process  is  continued  throughout  the  series  of  triangles  up  to  the  last  one,  where 
two  extra  moment  condition  Eqs.  (61q)  become  available  and  these  suffice  to  deter- 
mine the  three  first  assumed  values  K.  This  is  merely  a  process  of  successive  elimination, 
eminently  suited  to  the  present  problem. 

When  the  values  K  are  thus  found  then  the  several  moments  M  are  obtained  by  sub- 
stitution into  Eqs.  (61i).  See  the  example  below  for  symmetric  structures  with  symmetric 
loading. 

The  solution  of  the  whole  problem  is  possible,  because  for  any  frame  with  m  members 
and  p  panel  points,  there  will  be  £(ra  —  1)  triangles  furnishing  f(w  —  1)  equations  of  the 
form  of  Eqs.  (6lN,  o,  P)  and  p  equations  of  the  form  of  Eq.  (61q)  making  in  all  |(rw  —1)  +p 


ART.  Gl  SECONDARY  STRESSES  233 

available  equations  from  which  to  find  2m  unknown  moments  M.  Hence,  if  the  prob- 
lem is  determinate,  there  must  be  as  many  equations  as  there  are  unknowns  and 

2m  =  |(m-l)+p,  giving  2p=m+3 (61n) 

which  by  Eq.  (3x)  is  actually  true  for  any  determinate  frame  composed  of  triangles 
and,  therefore,  the  solution  is  possible. 

It  was  previously  pointed  out  that  the  assumptions  regarding  the  signs  of  the 
deflection  angles  r  as  illustrated  in  Fig.  6lD,  are  impossible,  and  that  not  all  these  angles, 
nor  the  moments  M  in  the  members  of  a  triangle  can  have  the  same  algebraic  sign  on 
account  of  the  condition  imposed  by  Eq.  (6lE).  Therefore,  the  computed  results  must 
furnish  deflection  angles  r  and  moments  M  of  both  signs. 

However,  to  avoid  complicated  rules,  it  is  necessary  to  adopt  some  conventional 
or  standard  figure  and  then  reverse  the  assumed  deflections  wherever  the  computed  angles 
T  are  negative.  For  this  reason  the  conventional  figure  of  distortion  is  assumed  as  indicated 
in  Fig.  61r,  wherein  the  members  in  alternate  triangles  I,  III,  "V,  etc.,  are  made  to 
present  a  concave  line  outward,  indicating  positive  moments  M. 


FIG.  61r. 

The  even  triangles  II,  IV,  VI,  etc.,  must  then  represent  negative  moments,  but 
in  order  that  Fig.  61r  may  represent  the  conventional  assumption  of  all  positive  moments, 
the  Eqs.  (3?D)  are  divided  through  by  minus  one  when  applied  to  the  even  triangles  with 
convex  lines  outward. 

Hence,  for  the  odd  triangles  with  positive  moments  M,  Eqs.  (3?D)  are  applied  in 

the  form 

E Ad  =  (fa  ~fb)   COt  r  +  (fa  ~fc)   COt  /?  1 

EJp  =  (fb-fc)  cota+(/6-/a)  cotr\, (61s) 

E AY  -  (fc  -fa)  cot  p  +  (fc  -fb)  cot  a  1 

and  for  the  even  triangles  the  negative  moments  are  also  made  positive  by  using  Eqs. 
(61s)  in  the  form 

=  (/6-/a)  cot  r  +  (fc-fa)  cot/?  I 
(fc-fb)  cota  +  (/«-/fc)  cotr     , (6lT) 

=  (fa  -fc)  cot  p  +  (fb  -fc)  cot  a 


234 


KINETIC  THEORY  OF  ENGINEERING  STRUCTURES      CHAP.  XI! I 


and  then  all  moments,  represented  by  Fig.  61r,  may  be  considered  positive.  The  sub- 
scripts a,  b,  and  c  in  Eqs.  (61s)  and  (Gli),  refer  to  the  sides  a,  b,  and  c  opposite  the 
respective  angles  a,  {3,  and  7-. 

Since  the  above  assumptions  do  not  represent  real  or  possible  conditions,  some  of 
the  moments  M,  resulting  from  the  final  computation,  must  develop  negative  signs. 
Thus,  if  Mac  and  M^  in  Fig.  61o,  should  turn  out  with  negative  signs,  then  the  real 
deformations  of  the  triangle  would  be  as  represented  in  Fig.  6 IG. 


FIG.  61n. 


In  evaluating  Eqs.  (61cj)  for  the  several  panel  points,  it  is  necessary  to  observe  the 
relative  signs  of  the  moments  in  the  several  members  according  to  the  following  direc- 
tions. Thus,  for  any  panel  point,  C,  Fig.  61n,  the  clockwise  moments  should  be  treated 
as  positive  and  the  counter  clockwise  moments  as  negative  to  obtain  the  following 
equation : 

=  +Moa-Mob+Moc-Mod=Q, 


or  in  terms  of  the  values  K 


+K 


6/0 

ion 


Qlob 

lob 


"    ° 


'On 


(61u) 


The  arithmetic  operations  required  in  the  solution  of  problems  are  somewhat  lengthy 
but  not  difficult  especially  when  the  values  7  and  I  of  the  members  are  well  rounded  off, 
which  is  always  permissible  in  these  computations. 

The  secondary  stress  resulting  from  the  bending  moments  M  =QKI/l,  produced  in  any 
member  by  the  riveted  connections,  may  now  be  found  from  Navier's  law,  Eq.  (49M),  as 


My_      QKy 


(61v) 


where  y  is  the  distance  from  the  gravity  axis  of  the  member  to  the  extreme  fiber  and 
fs  is  the  unit  secondary  stress  caused  by  the  bending  of  the  member  due  to  the  moment 
Mg,  or  Mb  applied  at  one  end  of  the  member. 

The  total  maximum  unit  stress  in  the  member  is  then  ±f±fs,  where  f=S/F,  which 
is  the  primary  unit  stress  due  to  the  direct  loading. 


ART.  01  SECONDARY  STRESSES  235 

Thus  for  a  compression  member,  the  maximum  total  stress  occurs  on  the  side  where 
fs  is  negative. 

The  eccentricities  e,  Fig.  6lA  or  6lB,  produced  at  either  end  of  any  member  ~AB  by 
the  secondary  moments,  may  be  found  from 


and      *!_.  ......     (61W) 


The  secondary  stresses  here  considered  may  vary  in  percentage  of  the  primary  stresses 
from  5  to  100  per  cent,  depending  on  the  type  of  truss  and  the  details  of  design  employed 
for  the  members. 

The  subject  is,  therefore,  one  of  considerable  importance  and  cannot  be  dismissed 
in  the  usual  manner  as  taken  care  of  by  the  safety  factor. 

The  question  of  maximum  secondary  stress  in  any  member  has  not  yet  received 
consideration  here,  though  it  is  a  matter  of  vital  importance. 

It  does  not  follow  that  the  maximum  secondary  stress  occurs"  simultaneously  with 
the  maximum  primary  stress  for  any  particular  member.  The  results  of  computations 
show  this  to  be  quite  true  for  the  chords,  but  not  for  the  web  members,  though  there 
seems  to  be  no  criterion  available  to  indicate  the  position  of  a  train  of  loads  to  produce 
maximum  secondary  stress  in  a  member. 

The  method  of  influence  lines,  while  not  impossible,  becomes  practically  prohibitive 
on  account  of  the  labors  involved.  The  only  reasonable  treatment  which  suggests  itself 
is  to  analyze  the  structure  first  for  total  maximum  load  and  then  for  a  few  typical  posi- 
tions for  the  moving  load  and  from  these  results  select  the  probable  maximum  values 
for  each  member. 

Since  the  entire  truss  must  be  completely  solved  for  each  simultaneous  position 
of  the  live  load,  this  appears  to  be  the  only  practicable  advice  to  give  in  view  of  the 
immense  labors  involved  in  such  analyses. 

The  above  method  is  now  illustrated  by  the  solution  of  a  problem. 

Example.  A  single-track,  through  Pratt  truss,  designed  by  Messrs.  Waddell  & 
Hedrick,  in  1899,  for  the  Vera  Cruz  and  Pacific  Ry.,  Mexico,  is  selected.  The  truss  has 
six  panels  with  a  span  of  145  feet,  a  depth  of  30  feet,  and  the  distance  between  trusses 
is  17  feet.  The  equivalent  uniform  live  load  for  WaddelTs  class  W  was  used.  The  case 
here  investigated  is  for  maximum  total  load  over  the  entire  span. 

The  cross-sections  F  and  the  moments  of  inertia  I  are  computed  and  tabulated 
together  with  the  lengths  I,  widths  b  and  distances  y  for  all  the  members  of  the  truss. 
These  together  with  the  stresses  S,  due  to  the  total  maximum  load  Fig.  61i,  are  given 
in  Table  61  A,  where  the  unit  stresses  f=S/F  and  the  functions  61/1  and  6y/l  are  com- 
puted. Impact  is  not  considered. 

A  live  load    of   54000   Ibs.   and  a  dead  load  20800  Ibs.  per  panel  were  assumed  for 

tke  computation  of  the  stresses. 

While  it  is  not  necessary  to  know  how  the  members  were  designed  as  long  as  F  and 
7  are  given  for    each,  yet  it  may  be  of  some  interest  to  the  reader  to  have  this 
Figs.  61  J  thus  represent  the  sections  of  the  several  members. 


236 


KINETIC  THEORY   OF  ENGINEERING  STRUCTURES      CHAP.  XIII 


FIG.  61i. 


B-D. 


D-F. 


A-C-E. 


E-G. 


B-C. 

T*8 


D-EANOF-G. 


B-EAnoD-G. 


II  1 

FIGS.  61J. 


C  E 

FIG.  6lK — Unit  Stresses  and  Cotangents 


ART.  61 


SECONDARY  STRESSES 


237 


TABLE  6lA 
DATA  FOR  THE  TRUSS  FIG.  61i. 


Mem. 

F 
Gross, 

Sq.in. 

7 

Gross, 

in.< 

/ 
in. 

6 

In. 

y 

In. 

s 

Lbs. 

f-S 
f-F 

Lbs.  sq.in. 

67 

l 

6v 
1 

AC 

15.9 

139 

290 

12.37 

6.19 

+  158,400 

+  9960 

2.88 

0.128 

CE 

15.9 

139 

290 

12.37 

6.19 

+  158,400 

+  9960 

2.88 

0.128 

EG 

26.9 

282 

290 

12.5 

6.25 

+  252,060 

+  9370 

5.84 

0.129 

BD 

26.5 

750 

290 

15.0 

7.5 

-252,060 

-9510 

15.52 

0.155 

DF 

29.4 

805 

290 

15.0 

7.5 

-283,500 

-9640 

16.66 

0.155 

AB 

33.5 

1464 

462 

16.25 

8.34 

-245,000 

-7310 

19.01 

0.108 

BE 

17.6 

323 

462 

12.0 

6.0 

+  156,200 

+  8875 

4.19 

0.077 

DG 

17.6 

323 

462 

12.0 

6.0 

+  48,400 

+  2750 

4.19 

0.077 

BC 

13.7 

119 

360 

12.37 

6.19 

+  68,000 

+  4960 

1.98 

0.103 

DE 

14.7 

288 

360 

12.0 

6.0 

-   44,200 

-3010 

4.80 

0.100 

FG 

14.7 

288 

360 

12.0 

6.0 

6,800 

-  460 

4.80 

0.100 

6  =  width  of  member  in  the  plane  of  the  truss. 

y  =  distance  from  neutral  axis  to  extreme  fiber  of  section. 

I  is  taken  about  the  gravity  axis  perpendicular  to  the  plane  of  the  truss. 

The  distortions  in  all  the  angles  of  the  several  truss  triangles  are  now  computed 
from  Eqs.  (61s)  and  Eqs.  (6lT),  using  the  unit  stresses/ from  Table  6lA  and  the  cotangents 
of  the  angles,  all  as  shown  in  Fig.  6lK. 

Applying  Eqs.  (61s)  to  the  odd  triangles  I,  III  and  V,  and  Eqs.  (6lT)  to  the 
even  triangles  II  and  IV  the  following  values  Ed  are  obtained. 

Triangle  I.     Eqs.  (61s) : 

EAai=(  4960-9960)0.0  +(  4960+7310)1.24  ==+15,215 
i=(  9960 +7310)0.806 +  (  9960-4960)0.0  ==+13,920 
i  =  ( -7310 -4960)  1.24  +(-73 10 -"9960)  0.806= -29,135 


00 


Triangle  II.     Eqs.  (6 IT): 

=  (     9960 -8875)0.806 +  (  4960- -8875)1.24   =-  3,980 

=  (     4960-9960)0.0     +(  8875-9960)0.806=-      875 

2  =  (     8875-4960)1.24   +(  9960-4960)0.0     =+4,855 


00 


238  KINETIC   THEORY    OF   ENGINEERING  STRUCTURES      CHAP.  XIII 

Triangle  III.     Eqs.  (61s): 

=  (  -3010  +9510)0.0  +(-3010-8875)1.24  =-14,737 
=  ( -9510-8875)0.806  +  ( -9510  +3010)0.0  =  - 14,818 
=  (  8875+3010)1.24  +(  8875+9510)0.806= +29,555 


00 
Triangle  IV.     Eqs.  (6  IT)  : 

=  (  9370 -2750)0.806 +  (-3010 -2750)  1.24  =-  1,806 
=  ( -3010  -9370)0.0  +  (  2750  -9370)0.806  =  -  .5,336 
=  (  2750+3010)1.24  +(  9370+3010)0.0  =+  7,142 


00 
Triangle  V.     Eqs.  (61s) : 

=  (-  460+9640)0.0  +(-  460-2750)1.24  --  3,980 
=  ( -9640  -2750) 0.806  +  (-9640+  460)0.0  =  -  9,986 
-(  2750+  460)1.24  +(  2750+9640)0.806= +13,966 


00 


The  condition  Eq.  (6lE)  must  be  satisfied  for  each  triangle  as  indicated  above,  offer- 
ing a  complete  check  on  the  numerical  results. 


FIG.  6lL — Values  EA. 

The  values  EA  are  now  entered  on  the  diagram  Fig.  6lL,  where  the  conventional 
assumptions  for  the  moments  M  are  indicated  according  to  Fig.  6 IF.  The  moment 
Eq.  (61q),  is  then  written  out  by  following  the  directions  given  under  Eq.  (61u). 

Using  the  values  for  67/7  from  Table  6lA  in  Eq.  (61q)  the  following  equations  are 
obtained : 


ART.  (51  SECONDARY  STRESSES  239 

For  panel  point  A, 


"ob  'oc 

and  after  inserting  the  values  from  Table  61  A,  this  gives 

19.01^=2.88^       or      Kac=Q.QQKab    .     ......     (1) 

For  panel  point  B, 

2  Mb  =  -  Mte  +  Mbc  -  Mbe  +  Mbd  =  0, 
giving 

-  ig.Oltffc,  +  1.98#6c  -4.  l9Kbe  +  15.52Kbd  =0, 
or 

Kbd  =  1.23^-0.12^  +0.27  Kbe  .........     (2) 

For  panel  point  C, 

2Mc  =  MCtt-Mc6+Mce=0, 
giving 

2.8&KM  -  1.98^  +2.8»KC9  =0, 
or 

^ca.       .     .     ........     (3) 


For  panel  point  D, 

2Md  =  -Mdb  +Mde  -Mdg  +Mdf  =0, 
giving 

-  15.52Kdb  +4.SOKde  -4.  l9Kdg  +  16.66  Kdf  =0, 

or 

(4) 


For  panel  point  E, 

SMe  =  -Mec  +Meb-Med+Men  =0, 

giving 

-2.88Kec  +4.19#e6  -4.8QKed  +5.84Keg  =0, 

or 


...     (5) 
For  panel  point  F,  noting  that  for  symmetrical  loading  Mfd=Mfh,  then 


panel  point  G,  where  for  symmetric  loading  AT,,  =--Mgi  and  Mgd  =  Mghi  then 


or 

-r  /A^ 

,•11? 71  r  Hf        . ,  ...  1     71/f ?IT  .      ff»on 

For 

giving  , ,  v       n 
-5MKge  +4.19Xffd  -4.80^/  +4.19A,,  -o.84^fft-  =0, 

#^--2.433^+1.746^. (7) 


240 


KINETIC  THEORY    OF  ENGINEERING    STRUCTURES       CHAP.XIJI 


When  there  is  no  middle  vertical  FG,  then,  for  symmetric  loading,  an  equation  of 
the  form  Eq.  (6lM)  is  written  out  for  the  center  triangle  DGH  to  supply  the  last  condition. 

For  an  unsymmetric  truss  or  unsymmetric  loading,  the  several  equations  must  be 
extended  over  all  pin  points  of  the  structure. 

Assuming  now  that  Kab,  K^  and  Kac,  Fig.  6lL,  are  known,  then  Eqs.  (6lN,  o,  P) 
will  furnish  the  values  K^,  Kbc  and  Kcb  for  the  first  triangle  ABC,  and  by  applying  this 
process  successively  to  all  the  triangles  it  becomes  possible  to  express  all  the  values  K 
in  terms  of  Kab  and  K^  by  including  the  moment  equations  (1)  to  (5)  just  evaluated. 
The  two  last  Eqs.  (6)  and  (7)  will  then  serve  to  find  Kab  and  K^,  as  will  be  shown  later., 
and  these  values  substituted  back  will  furnish  all  the  unknowns  K. 

In  applying  Eqs.  (6lN,  o,  P)  to  the  several  triangles,  it  is  decessary  to  use  the  standard 
lettering  employed  in  the  derivation  of  these  formulae  and  hence  each  triangle  is  sketched 
in  Figs.  6lM,  to  exemplify  the  process.  The  angle  a  must  be  so  located  in  a  triangle 
that  it  is  included  between  the  two  adjacent  known  K's,  and  the  angle  /?  must  be  adjacent 
to  the  third  known  K,  while  f  is  between  the  adjacent  unknown  K's.  This  then  deter- 
mines the  vertices  (A),  (B)  and  (<?),  shown  in  parenthesis,  for  the  standard  lettering. 


FIGS.  6lM. 

The  designations  of  the  angles  in  Fig.  6lK,  for  the  solution  of  the  values  EA  were 
selected  so  as  to  comply  with  the  conditions  just  described.  The  arrows  in  the  Figs. 
6lM  indicate  the  values  K  found  from  a  previous  triangle  and  from  one  new  moment 
Eq.  (6lQ)  above  evaluated  as  (1)  to  (5). 

Eqs.  (6lN,  o,  P)  and  Eqs.  (1)  to  (5)  thus  give: 

At  panel  point  A,  by  Eq.  (1) :     Kac=Q.QKab. 


By  Eq.  (6lN),  KM 

By  Eq.  (61o),  Kbe 

By  Eq.  (61p),  KA=EAn+Kta>+Kl>a-Kea^  -44,350  +  16.2^ 

At  panel  point  C,  by  Eq.  (3),  Kee=O.Q9KA-KM=  -45,816+26.4^+2.38^. 

I  By  Eq.  (6lN),  Kec=EJa2 -2Kee-2KA -#6c  =  147,2 17 -76.6^-6.76^ 

II  \  By  Eq.  (61o),  Kbc=E4p2+Kcc-Khc+Kec-=71,3Ql  -41.6^-2.38^ 

By  Eq.  (61p),  Keb=EAr2+Kcb+Kbc-Kec=  - 157,577 +84.2#a6 +6.767^. 


ART-6l  SECONDARY  STRESSES  241 

At  panel  point  B,  by  Eq.  (2) :  ^M  =  l-23Xte-0.12X6c+0.27X6e  =  15,780-10.2Xo6+0.83X6([. 

[  ByEq.  (61N),    Kdb=EJa3-2Kbd-2Kbe-Keb=  -31,502  +  19.4Xa6-3.66X6o 
AIII  j  ByEq.  (61o),    Ked~EJi83+KM-Keb+Kdb  =  l 27,037 -75.0Xa6-9.59Xj>a 

I  By  Eq.  (61p),    Kde=E4r3+Kbe+Keb-Kdb=  -25,129  +23.2Xa6+8.04X6a. 

At  panel  point  E,  by  Eq.  (5) : 

Keg  =0.493Xec  -0.72Xe6  +0.82Xed  =290,203  -  159.9Xn6  -  16.07X6a. 

I"  By  Eq.  (6lN),    Kge  =EJa^-2Keg-2K(id-Kde=  -811,157+446.6Xa6+43.28X6a 
AIV  I  ByEq.  (61o),    Kdg=EJl34+Keg-Kde+Kge=  -501,161  +263.5X^  +  19.17X6,, 

I  By  Eq.  (61p),    Kgd  =E4r4+Ked+Kde-Kge  =920,207  -498.4Xtt6-44.83X6a. 

An  panel  point  D,  by  Eq.  (4) : 

Kd1  =0.932Xd6  -0.288Xde  +0.252Xdff  =  -148,414  +77.80  Kab  -O.gOX^. 
ByEq.  (6lN),    Kfd  =EAa5-2Kdf-2Kdg-Kgd  =374,963  -184.2Xa6+8.29X6a 
AV     ByEq.  (61o),    Kgf=E^s  +  Kdf-Kad  +  Kfd=  -703,644  +392.0Xab+52.22Xfca 


By  Eq.  (61?) ,    Kfa  =EJr5  +Kdg+Kgd -Kfd  =58,049 -SOJX^ -33.95^. 

The  moment  Eqs.  (6)  and  (7)  for  panel  points  F  and  G  now  furnish  independent 
values  for  X/ff  and  Kgf  and  by  combining  these  with  the  last  two  equations  of  AV,  the 
values  Kab  and  X^  may  be  found  as  follows: 

By  Eq.  (6),  X/ff=6.942X/d  =2,602,993 -1278.7Xa6+57.55X6a. 

By  Eq.  (7),  ^--2.433^+1.746^-3,580,226-1056.81^-183.57^. 

Substituting  these  values  of  Kfg  and  Kgj  into  the  last  two  equations  of  AV,  then, 

2, 544,954 -1228.0Xaft+  91.50X^=0, 
4,283,870  -2348.8Xa6  -235.79Xfea  =0, 

from  which  Kab  =  1947.5  and  Kba=  -1676.8. 

These  values  of  Kab  and  X^  are  now  substituted  into  the  above  twenty  equations, 
furnishing  all  the  values  X  as  follows: 

(1)  Kab  =  1947.5 

(2)  Kba=- 1676.8 

(3)  Xac=6.6Xab  =  12,854 

(4)  #„  =  15,215-  29,602+  1,677= -12,710 


242  KINETIC  THEORY  OF   ENGINEERING   STRUCTURES       CHAP.  XIII 

(5)  Kbc=       29,135-    16,748+  3,354=      15,741 

(6)  Kcb=-    44,350+  31,549-  3,354  = -16,155 

(7)  Kcc=-   45,816+  51,414-  3,991=        1,607 

(8)  Kec=     147,217-149,178  +  11,335=       9,374 

(9)  Kbe=       71,391-  81,016+  3,991=    -  5,634 

(10)  Keb=- 157, 577  +  163 ,979  -11, 335=-    4,933 

(11)  Kbd=       15,780-19,864-   1,392-   -  5,476 

(12)  Kdb=--  31,502+  37,781+  6,137=     12,416 

(13)  Ked=     127,037-146,062  +  16,081  =  -  2,944 

(14)  Kde=-  25,129+  45,182-13,481=       6,572 

(15)  Keg=     290,203-311,405+26,946=       5,744 

(16)  Kge  =  -811,157  +869,753  -72,572  =  -13,976 

(17)  Xd,= -501,161 +513,166 -32,144  =  -20,139 

(18)  Kgd=     920,207-970,634+75,171=      24,744 

(19)  Kdf=- 148,414  +  151,515+   1,509=       4,610 

(20)  Kid=     374,963-358,730-13,901=        2,332 

(21)  Kgf=  -703,644  +763,420  -87,562  =•  -27,786 

(22)  Kta=       58,049-  98,738+56,927=     16,238 

Eq.  (6lM)  now  furnishes  a  convenient  check  on  the  above  numerical  results,  by  tak- 
ing the  sum  of  all  the  six  values  K  for  each  triangle. 

Thus  for  AI,  the  values  (1)  to  (6)  give 

S#i  =30,542 -30,542  =  0 
and  for  All,  the  values  (5)  to  (10)  give 

2K2=  26,722 -26,722=  0 
and  for  AIII,  the  values  (9)  to  (14)  give 

2#3  =  18,988 -18,987=  ?. 
and  for  AIV,  the  values  (13)  to  (18)  give 

2  A'4  =37,060  -37,059  =  1 
and  for  AV,  the  values  (17)  to  (22)  give 

2^5=47,924-47,925=  -1 

The  secondary  stresses  due  to  the  bending  at  each  end  of  each  member  are  now  found 
from  Eq.  (61v)  as  given  in  the  following  Table  6lB. 


ART.  (51 


SECONDARY  STRESSES 


243 


TABLE  6lB 
SECONDARY  STRESSES  DUE  TO  BENDING 


Mem. 

End. 

6y 
I 
in. 

K 

±/.-5? 

Eq.  (61v). 
Lbs.  sq.in. 

'-£ 

Lbs. 
sq.in. 

67  K 

"   SI 
in. 

Mem. 

End. 

E 
D 
E 
G 
D 
G 
D 
F 
G 
F 

6.V 
I 
in. 

K 

±'.=6JP 
Eq.  (61y), 
Lbs.  sq.in. 

/-I 

Lbs. 
sq.  in. 

61  K 

~  SI 
in. 

AB 

AC 
EC 
CE 
BE 
BD 

A 
B 
A 

C 
B 
C 
C 
E 
B 
E 
B 
D 

0.103 

+   1,948 
-    1,677 
+  12,854 
-12,710 
+  15,741 
-16,155 
+   1,607 
+   9,374 
-   5,634 
-  4,933 
-   5.476 
+  12,416 

200 
172 
1645 
1627 
1621 
1664 
205 
1200 
434 
380 
849 
1924 

-7310 
-7310 
+  9960 
+  9960 
+  4960 
+  4960 
+  9960 
+  9960 
+  8875 
+  8875 
-9510 
-9510 

0.15 
0.13 
0.23 
0.23 
0.46 
0.47 
0.03 
0.17 
0.15 
0.13 
0.34 
0.72 

ED 
EG 
DG 
DF 
GF 

0.100 

-   2,944 
+   6,572 
+   5,744 
-13,976 
-20,139 
+  24,744 
+   4,610 
+   2,332 
-27,786 
+  16,238 

294 
657 
741 
1803 
1551 
1905 
715 
361 
2779 
1624 

-3010 
-3010 
+  9370 
+  9370 
+  2750 
+  2750 
-9640 
-9640 
-   460 
-   460 

0.32 
0.72 
0.13 
0.32 
1.76 
2.16 
0.28 
0.14 
19.62 
11.46 

0.128 

0.129 

0.103 

0.077 

0.128 

0.  155 

0.077 

0.100 

0.155 

The  total  stress  on  the  extreme  fiber  is  fs  +/,  noting  that   no  increase  was  made  in  /  for  buckling 
effect  in  compression  members. 

The  actual  signs  of  the  values  K,  and  hence  also  of  the  moments  M^&KI/l,  now  being 
determined,  the  real  character  of  the  distortions  may  be  represented  diagrammatically 
in  the  following  Fig.  6 IN. 


For  compression  members  the  most  severely  stressed  fibers  will  occur  on  the  side 
where  /.  is  negative  and  for  tension  members  on  the  side  where  /„  is  positive.     Thus  in 
the  compression  member  AB,  the  critical  stressesoccur  on  the  upper  side  at  A  and  on  the 
lower  side  at  B,  while  for  the  tension  member  AC,  these  occur  on  the  upper 
and  on  the  lower  side  at  C. 


244  KINETIC  THEORY   OF   ENGINEERING  STRUCTURES      CHAP.  XIII 


ART.  62.     SECONDARY   STRESSES   IN   RIVETED   CROSS   FRAMES   OF   TRUSSES 

The  analysis  of  cross  frames,  so  far  as  this  is  possible,  presents  many  obstacles,  some 
of  which  cannot  be  overcome  owing  to  the  variable  character  of  the  external  loads. 

The  bending  effect  due  to  all  forces  acting  on  a  frame,  even  when  direct  stresses  are 
neglected,  leads  to  very  complicated  formulae  of  more  or  less  questionable  value,  and 
no  attempt  is  made  here  to  discuss  the  general  problem. 

A  cross  frame  is  usually  subjected  to  the  dead  load  of  the  bridge  floor  concentrated 
on  the  floor  beam;  the  live  load,  impact  and  centrifugal  force  applied  to  the  floor;  wind 
pressure  against  the  moving  load  and  the  vertical  trusses;  and  unequal  deflection  of  the 
main  trusses  due  to  a  variety  of  causes,  but  particularly  to  unequal  temperature. 

The  magnitudes  of  the  secondary  stresses  depend  of  course  on  the  details  of  construc- 
tion and  bracing  employed  in  any  special  frame,  so  that  many  different  forms  would 
require  investigation.  However,  only  two  principal  types  will  be  discussed  and  the 
same  formulae  are  applicable  to  both  through  and  deck  bridges. 

In  the  following,  the  unbraced  and  one  type  of  braced  cross  frames  are  considered 
first  for  direct  loading  and  then  for  wind  effect. 

(a)  Dead  and  live  load  effects.  Unbraced  cross  frame  with  rigid  post  connections.  The 
construction,  shown  in  Fig.  62A  with  lettered  dimensions,  is  analyzed  by  assuming 
equality  between  the  unknown  bending  moments  induced  in  the  posts  by  the  symmetric 
loads  P,  as  follows:  M0  =  M2  and  M]  =MS,  positive  as  indicated  when  they  produce 
compression  on  the  outer  fibers  of  the  posts. 

Call  M  the  bending  moment  on  the  floor  beam  due  to  symmetrically  placed  loads 
P  acting  on  a  simple  beam  which  rests  on  two  supports  C  and  D. 

Also  call  mi  the  moment  at  any  point  of  the  post  AC  distant  x  from  A  ;  m?.  the 
moment  at  any  point  of  the  strut  AB;  and  m^  the  moment  at  any  point  of  the  beam 
CD.  Then  from  Fig.  62A. 

ml=M1+(M°~Ml\x,      m2=M1       and      m3=M0+M.    .     .     .     (62A) 
\       h       / 

Considering  only  the  effect  due  to  bending,  by  neglecting  shear  and  direct  thrust, 
the  virtual  work  of  deformation  for  the  entire  frame  would  be  by  Eq.  (15n) 


~  Cmfdz.    .-   .     :     .     .     .     .     .     .     .     .     (62n) 

-&/3./0 


~ 

-&/3 

Substituting  the  values  from  Eqs.   (62A)  into  Eq.  (62p,)  then 


which  integrated  by  considering  everything  constant  except  x,  gives 

?^.+   *   \M0*b+2M0  fbMdx+  P^WI,     (62c) 

£/2         bl3\_  J®  Jo 


ART.  62 


SECONDARY  STRESSES 


245 


which  is  the  general  expression  for  any  case  of  loading  where  M  must  be  evaluated  for 
each  case. 

Now  since  the  unknown  moments  M0  and  MI  must  have  such  values  as  will  make 
the  first  differential  derivative  of  W  with  respect  to  each,  equal  to  zero,  then  after  some 
reduction 

'dW 


and 


3M0      3/i 


_  ^  =f) 

- "  " 


(62n) 


?•*"    m*     II 

•-j 

,-"" 

"* 

.-I,          I,- 

T 

50         ITIj       Ij         - 

D 

FIG.  62A. 

FIG.  62B. 


Solving  Eqs.  (62n)  gives 


,.    6l\o  .  0\     oil 

=  -Mn(-r^-+2     -TJ- 


and 


from  which 


(62s) 


Mi  = 


^i-1       Mdx 


3/i6 


(62p) 


-1 


1/2     '-/\W3 

For  the  symmetric  loading  shown  in  Fig.  62A,  the  integral  Mdx  becomes 


and  for  a  uniform  load  p  per  foot  of  floor  beam, 

fb  f)b3 

JQ 


(62G) 


246  KINETIC  THEORY  OF  ENGINEERING  STRUCTURES       CHAP.  XIII 

MQ  and  MI  may  thus  be  found  from  Eqs.  (62F)  and  (62r),  and  from  these  the  moments 
TOI,  m-2  and  ra3,  for  any  point  of  the  frame,  are  given  by  Eqs.  (62  A). 

The  maximum  fiber  stress  in  the  posts  must  be  combined  with  the  direct  stress  sus- 
tained as  a  truss  member. 

The  upper  strut  receives  a  compression  of  (Mo—MJ/h  and  a  bending  moment  MI 
over  its  entire  length. 

The  floor  beam  will  receive  a  bending  moment  of  Mo+M  and  a  direct  stress  of 
(Mi-Mo)  /h. 

Example.  Given  the  cross  frame  at  DE,  Fig.  61i,  with  the  following  dimensions: 
h  =28  ft.,  b  -=17  ft.,  a  =  7  ft.,  P  -59000  Ibs.,  h  =226  in.4  =  0.0109  ft.4,  72  =  1134  in.4  =0.0547 
ft.,4  and  73  =  15262  in.4  =0.736  ft.4  The  style  of  the  frame  is  as  shown  in  Fig.  62A. 

Using  the  value  from  Eq.  (62c)  in  Eq.  (6.2r),  and  substituting  the  above  data,  then 

3X00109     59000 
__  28X0.736*     4     (U       n  __ 

/3X0.0109X17       W3X0.0109X17       \  -—    > 

V   28X0.0549          )\     28X0.736  / 

and  from  Eq.  (62s) 


For  t/=5.5  inches,  Eq.  (61v)  then  gives  the  secondary  stress  in  the  post  DE. 

t      ,,  ,Miy     1499X12X5.5 

for  the  upper  end  D,  fd  =  ±-^~  =—  —  =438  Ibs.  per  sq.in.; 

/I 


t      .,     .  ,M0v    3541X12X5.5 

for  the  lower  end  E,  fe  =  ±—  p-  =—  —  =  1034  Ibs.  per  sq.m. 

1\  22u 


The  actual  deformation  of  this  frame  is  indicated  in  Fig.  62s. 

A  braced  cross  frame  with  rigid  post  connections  Fig.  62c,  will  now  be  treated  as  in 
the  previous  case. 

Neglecting  shear  and  direct  stress  as  before,  and  dealing  only  with  the  effect  due 
to  bending,  then  Eq.  (62B)  becomes 


Integrating  this  expression  as  above,  and  then  differentiating  first  with  respect 
to  MQ  and  then  with  respect  to  MI  and  placing  these  derivatives  equal  to  zero,  the  follow- 
ing equations  are  obtained: 


~~  =  A!  M0  +  2hi  Ml  +  2eMi  -  0. 


ART.  62  SECONDARY  STRESSES 

Solving  these  two  zero  equations  and  noting  that  h=h\+e,  then 


247 


i  M0 


and 


f 

Jo 


Mdx 


(62n) 


/*6 

wherein  I    Mcfo  is  given  by  one    of   Eqs.  (62a)  and   the   moments   MQ  and   M\    may 

Jo 

thus  be  found. 

The  stresses  in  the  upper  struts  then  become 


MI  M, 

=  --         and       /Se/  =  --  1 
e  e 


t. 


*, 


I. 


B 

11 


/• 

»H 


1L. 


-m, 
-I. 


K 


>•   • .. •  —  _ 

FIG.  62c.  FIG.  620. 

and  the  stresses  in  the  post  will  be 

and 


FIG.  62E. 


Example.     Taking  dimensions  as  in  the  previous  example  and  making   hi  =19.5  ft. 
then  M0  =4770  ft.-lbs.,  and  ^=1661  ft.-lbs.  giving  /,  =  1393  Ibs.  per  sq.in.,  and  fc  = 

Ibs.  per  sq.in.  . 

(b)   Wind  effects.     Unbraced  cross  frame  with  rigid  post  connections. 

wind  loads  on  the  trusses  of  a  bridge  may  be  carried  to  the  abutments  by  means  of  com- 
plete horizontal  trusses  in  the  planes  of  the  top  and  bottom  chords  respectively,  pro 
vided  the  end  reactions  of  the  top  chord  wind  system  can  be  carried  to  the  abutments 
by  suitable  end  postal  bracing.     In  this  case  the  intermediate  cross  frames  suff< 
little  or  no  distortion  and  hence  carry  no  bending  effects. 

However   when  the  wind  pressure  along  the  top  chord  is  carried  down  to  the  I 
torn  chord  locally  at  each  cross  frame,  then  the  latter  must  be  distorted  and  thus  resi 
bending      In  this  case  the  total  wind  pressure  is  carried  to  the  abutments  through  the 
bottom  chord  lateral  system.     The  external  forces  on  a  cross  frame  are  then  as  repn 


248 


KINETIC  THEORY  OF  ENGINEERING  STRUCTURES       CHAP.  XIII 


sented  in  Fig.  62D.     The  wind  loads  w  are  transmitted  through  the  posts  to  the  lower 
lateral  system,  producing  bending  effects  as  shown  in  Fig.  62E. 

Assuming  that  the  wind  loads  w  are  equal  and  calling  T  the  tangential  stress  at  the 
point  of  counterflexure  and  S  the  normal  stress  at  the  same  point,  then 


H=w  = 


(62i) 


The  general  work  equation  for  the  frame,  considering  bending  effects  only,  is  the 
same  as  Eq.  (62fi)  in  which  the  moments  mi,  m2  and  ra3  must  be  evaluated  for  the  wind 
forces  now  acting. 

The  moments  of  the  external  forces  about  A,  B,  C,  D  and  0,  from  Fig.  62o  (count- 
ing clockwise  moments  positive)  are  respectively 


M3=-  T(h  -ho)  -Sb  =H(h  ~ho)  =  - 


(62J) 


M2  =  ThQ  —Sb  —2wh  =  —Hh0  because 
-Sb-2w(h-ho)=0 
The  moments  mi,  m2  and  m^  thus  become 

mi=Mi  +  (  —  ^-  —  -  J  x=H(x  +ho—h)  with  origin  at  A 
\       h       I 


(  —  ^-r 


-\x=H(h—ho)(-r—l}  with  origin  at  A 


~with  origin  at  C 


(62K) 


Substituting  these  values  into  Eq.  (62s)  then 
2#2  Ch, 

W=ETJO  (x+h°~v 

which  gives  by  integration, 


-     dx 


(62L) 


Since  &o  must  be  so  chosen  that  the  internal  work  will  be  a  minimum,  this  particular 
value  may  be  obtained  by  equating  to  zero  the  differential  derivative  of  W  with  respect 
to  ho,  thus  obtaining 

^W     h  SRI      QM  _i_  b  (i>      i  \  j_^°&     n 
5r-  =7-(6rto— 3ft)  +-r(h0-hi)  +-j-  =0, 

from  which 


b_     Oh     b' 

/3       /I        /2 


(62M) 


ART.  62 


249 


Substituting  this  value  of  ho  into  Eqs.  (62X),  will  give  the  moments  at  all  points 
of  the  frame  and  from  Eqs.  (62j)  the  stress  S  and  the  moments  at  the  ends  of  the  post 
may  be  found. 

Example.  For  the  cross  frame  of  the  first  example  and  a  horizontal  wind  load  of 
Ibs.  the  following  values  are  obtained: 


oo 
2S 


3X28 
(XOl09 


0.0547  / 


17        6X28          17_ 
0.736      0.0109     0.0547 


- 14.25  ft. 


Substituting  values  into  Eqs.  (62j),  then, 

Mi  =  -H(h-ho)  =  -3100(28-14.25)  =  -42,625  ft.-lbs. 
MO  =Hho  =3100  X  14.25  -44,175  ft.-lbs. 
_     2w(h  -ho)  _      6200(28  - 14.25) 


b  17 

This  gives  for  the  stress  at  the  bottom  of  the  post  AC, 


=  -5197  Ibs. 


M0y_S 
'  /i       F 


44175X12X5.5     5197       ,100ftn     0.0  „ 

~T4~7  =  ±  12,899  — 3o3  Ibs.  per  sq.m. 


226 


For  a  braced  cross  frame  with  rigid  post  connections  Fig.    62r,  the  analysis   is   con- 
ducted in  precisely  the  same  manner  as  in  the  previous  case,  making  as  before,  H  =w  =  T. 

/t 


Counting  clockwise  moments  positive,  then  the  moments  of  the  external  forces  about 
the  points  E,  C  and  0  are  respectively 

Ml  =  -T(hl-ho)  =  -H(hl-ho)    ' 

M0  =  Tho=Hho  ....    , -.  ..    .    .    (62N) 

-Sb-2w(h-ho)=0 


250  KINETIC  THEORY  OF  ENGINEERING  STRUCTURES       CHAP.  XIII 

The  moments  at  any  point  x  from  C  then  become: 

for  Cl,  Mo  +  (Ml^M-}x  =  H&o  -*) , 

i,\ 

(62o) 


for  EA,  Mi  —(x  -h,)  =H(hl  -h0)  (—-}, 

e  \     e    / 

for  CD,  Mo  ~»x  =HhJl  ~   . 


These  values  inserted  into  the  general  work  Eq.  (62B)  give  for  the  frame 


i     . 
---     dx+-- 

e 


from  which  is  obtained  after  integrating 

i    *e^^H 
*o)  q-    +  Q 

o  J         o 


Taking  the  first  differential  derivative  with  respect  to  /?0  and  equating  the  same 
to  zero  then 


3fto     ^i 
from  which 


MM 
0=  -     ..............     ^      ^ 


Example.  Given  a  cross  frame  as  per  Fig.  62r,  for  which  ft  =28  ft.,  7^  =19.  5  ft., 
6  =  17  ft.,  e=8.5  ft.,  /i  =226  in.4,  73=  15262  in.4  and  w>=3100  Ibs.  to  find  /%  M0  and  Ml. 
From  Eq.  (62p) 

19.5(56  +  19.5) 

ho=  -  iT><226  =  10-96ft-' 
2(28+39)+^ 

and  from  Eqs.  (62N) 

MI  =  -3100(19.5-10.96)  =  -26,474  ft.-lbs. 
M0  =3100X19.96  =33,976  ft.-lbs. 

^=-^^0)  =  -6215  Ibs. 
6 

The  maximum  stress  in  the  post  ~AC  is  then 


±M«y  _S^  _     88976X12  XBJ  _^18  _±9928  _423  Ibs.  per  sq.in. 
/i        /'ac  226  14.7 

The  stresses  in  the  top  struts  and  diagonal  are  not  so  easily  determined  when  the 
double  latticed  type  of  bracing  is  used.     In  the  present  case  with  one  diagonal  AF,  the 


ART.  03  SECONDARY  STRESSES  251 

stresses  are  determined  by  passing  a  section  tt  through  the  three  members  and  the  point 
of  counterflexure,  and  taking  moments  about  A,  then, 

P-^T        3100(28-10.96) 

_T(h—hv)  -eEF=Q,      or      EF=-  —        — ^—      -  =  -62i5  Ibs. 

o.o 

The  stress  in  AJ3  is  found  by  taking  moments  about  F,  obtaining, 

- T(hi  -h0)  -Sb -we  +eAB  =0, 
whence 

T(hl  _^)  +Sb  +we     3100(19.5 -10.95) -6215X17 +3100X8.5  = 
AB=-  —T  8.5 

Also  from  the  vertical  shear  on  the  section 

ji  a  =0,       or      TF  =  --^-  =  +^a  =  !3,842  Ibs. 


In  concluding  the  subject  of  cross  frames  it  might  be  added  that  good  designs  should 
aim  at  deep  floor  beams  and  slender  vertical  posts,  as  is  clearly  indicated  by  Eq.  (62n)  , 
which  shows  that  M0  is  diminished  when  73  is  increased  and  increased  when  Ir  is  increased. 

It  has  been  stated  by  Mr.  Grimm,  "  Secondary  Stresses  in  Bridge  Trusses,"  p.  80, 
that  the  assumption  of  fixed  connections  between  floor  beams  and  posts  is  not  verified 
by  investigations.  The  author  suggests  an  explanation  for  this  by  calling  attention 
to  the  fact  that  recent  tests  indicate  that  compression  members  as  formerly  built  are 
not  nearly  as  stiff  against  buckling  as  was  supposed  and  furthermore,  a  slight  initial 
deformation  approaching  the  elastic  curve  which  the  stressed  member  might  attain, 
would  almost  obviate  secondary  stresses  in  the  cross  frames.  That  such  deformations 
actually  exist,  or  might  be  produced  in  overloaded  posts  by  a  permanent  set,  there  can 
be  little  doubt,  and  hence  it  is  quite  easy  to  understand  why  some  of  the  high  theoretical 
stresses  do  not  appear  to  exist  when  the  actual  stresses  are  measured  with  mstrum 

ART.  63.     SECONDARY   STRESSES   DUE   TO   VARIOUS   CAUSES 

(a)  Bending  stresses  in  the  members  due  to  their  own  weight.  The  maximum  bending 
moment,  in  any  member,  resulting  from  its  own  weight,  when  supported  on  fnctionless 
pin  bearings  at  the  ends,  will  occur  at  the  center  of  the  member  and  i 


(63A) 


where  I  is  the  length  of  the  member,  p  its  weight  per  unit  of  length  and  6  is  the  angle 
which  the  member  makes  with  the  horizontal. 

WhenThe  member  is  fixed  at  the  ends  and  there  is  no  direct  stress  the  bendmg  moments 
M0  at  the  ends  of  the  member,  and  Mc  at  the  center,  become  respe, 


and 


252  KINETIC  THEORY  OF  ENGINEERING   STRUCTURES      CHAP.  XIII 

In  any  case  these  moments  produce  a  unit  stress  /  on  the  extreme  fibers  of  the  mem- 
bers which  is  given  by  Navier's  law,  Eq.  (49M)  as 


(630 


where  y  is  the  normal  distance  from  the  neutral  axis  to  the  extreme  fiber. 

.  It  is  thus  seen  that  M  increases  as  the  square  of  I  while  /  is  inversely  proportional 
to  7.  Hence,  long  members  of  small  moment  of  inertia  may  receive  severe  secondary 
stress  due  to  their  own  weight. 

When  the  member  is  in  compression  or  tension,  the  combined  bending  stress  on  the 
extreme  fiber,  due  to  the  direct  stress  and  the  uniform  load,  must  be  investigated.  This 
cannot  be  accomplished  by  algebraic  summation  of  the  separate  bending  effects,  because 
axial  compression  increases  the  deflection  due  to  cross  binding,  while  axial  tension  dimin- 
ishes this  deflection.  The  bending  stress,  resulting  from  the  simultaneous  loading  must, 
therefore,  be  found. 

The  analysis  for  riveted  end  connections,  while  very  complicated,  is  not  usually 
necessary.  For  pin-connected  members  where  the  cross  bending  effect  is  always  more 
severe,  the  following  approximate  solution  is  given. 

A  horizontal  compression  member,  with  centric  pin  connection  at  each  end,  is  shown 
in  Fig.  63A.  Let  Mc  designate  the  maximum  bending  moment  which,  in  this  case,  occurs 
at  the  center  of  the  member.  Calling  p  the  weight  of  the  member  per  unit  of  length, 
S  the  axial  stress,  I  the  length  of  the  member  and  3  the  deflection  at  the  center,  then 

I   ,'.'.-..    .    .     (63D) 


_=_         8Sd 
~48ET~3MEIP 
from  which 


Assuming  the  effect  of  the  direct  stress  on  the  deflection  d  to  be  the  same  in  character 
as  that  of  the  uniform  load  p,  which  is  not  quite  true,  then 


(63E) 


(63F) 


48EI -5S12' 
This  value  of  d  substituted  into  Eq.  (63D)  gives 

6pm 


ART.  63  SECONDARY  STRESSES  253 

The  combined  direct  and  bending  unit  stress  on  the  extreme  fiber  of  the  member 
then  becomes  by  Eq.  (49M) 


(030) 


where  F  is  the  cross-section  and  y  is  the  normal  distance  from  the  neutral  axis  to  the 
extreme  fiber  of  the  member. 

Horizontal  tension  members  with    pin-connected   ends,  when    similarly  treated,  lead 
to  the  following  formulae: 


,      „         6pPE7 
"     •"*      M<=48W+» (63H) 


The  combined  direct  and  bending  stress  on  the  extreme  fiber  is  again  found  by 

Eq.  (63o). 

Example-  Given  an  eye  bar  2 X 15  in.,  making  F  =30  sq.in.;  5=600,000  Ibs.;  Z=660 
in.;  7=562.5  in.4;  p=8.5  Ibs.  per  in.;  and  E=  29, 000 ,000  Ibs.  per  sq.in. 

Then  from  Eqs.  (63n),  3  =0.4824  in.;  and  Me--=  173 ,450  in.  Ibs.  The  stress  on  the 
extreme  fiber  then  becomes  by  Eq.  (63c) 

,600,000    173,450X7.5  =±2313  lbg  . 

J         30  562.5 

showing  that  the  bending  stress  alone   increases  the  tension  in  the  lower  fiber  by  11.5 

per  cent. 

(b)  Secondary  stresses  in  pin-connected  structures.     In  all  usual  stress  computations 
the  question  of  how  equilibrium  is  established  between  the  stresses  in  the  several  mem- 
bers meeting  at  a  pin  or  panel  point  is  not  considered.     It  is  thus  assumed  that    the 
individual  members  extend  to  the  pin  centers  with  full  effective  sections  where  the  stres 
are  suddenly  balanced  in  a  point. 

In  reality  the  case  is  very  different  and  no  such  balance  in  the  stresses  can  be  accom- 
plished The  nearest  approach  to  a  realization  of  the  ideal  condition  is  in  the  case  of  two 
abutting  compression  members  and  then  only  when  there  are  no  bending  stres 

be  transmitted.  ,,     ,      , 

When  all  the  members  are  connected  by  pins  according  to  the  usual  methods  of 
construction,  the  stresses  are  transmitted  past  the  panel  points  in  very  much  the  same 
manner  as  for  riveted  connections  on  account  of  the  factional  resistance  on  the  pins 
created  when  the  structure  is  distorted  by  superimposed  loads^  The  advantages  uu^ 
claimed  for  pin  connections  are  not  all  realized  in  pract.ce  and  wh,  e  cent™  connections 
are  best  made  by  this  stvle  of  panel  joint  there  may  be  suffic.ent  fnctional  resistance  on 
the  pin  to  produce  secondary  bending  stresses  qnite  equal  to  those  occurring  m  rive. 

is  thus  an  erroneous  idea  to  regard  a  pin-connected  structure  as  free  from  secondary 


254  KINETIC  THEORY  OF  ENGINEERING  STRUCTURES        CHAP.  XIII 

stresses.     This  is  never  so  and  the  rnalysis  may  be  even  more  complicated,  though  the 
results  are  probably  more  reliable  than  for  riveted  structures. 

The  exact  behavior  of  a  member  with  pin  bearings  depends  on  many  circumstances, 
especially  the  diameters  of  the  pins.  When  these  are  sufficiently  large  the  member  can- 
not turn  under  load,  and  secondary  stresses  must  result.  These  may  be  partially  relieved 
by  the  vibratory  action  of  the  live  loads  which  would  probably  allow  the  structure  to 
adjust  itself  for  some  mean  condition  of  loading  and  reduce  the  secondary  stresses  to  a 
minimum.  However,  temperature  changes  would  enter  as  a  disturbing  element  to  pre- 
vent such  favorable  action  from  establishing  permanent  relief. 

The  frictional  resistance  offered  by  a  pin  connection  is  now  analyzed. 
Let  S  =the  stress  in  any  member. 

R  =the  total    frictional  resistance  on  the  surface  of    contact  between  the  pin 

and  the  member. 
r=the  radius  of  the  pin. 
^o— the  angle  of   repose,  making  tan  ^o  =  (a=the  coefficient  of   friction  between 

the  two  metals. 

c  =the  eccentricity  of  S  required  to  resist  the  moment  of  the  frictional  force  R. 
e=the  actual  eccentricity  of  S  required  to  resist  the  bending  moment  Ma  due 

to  secondary  stress. 

The  line  of  stress  S,  Fig.  63s,  is  supposed  to  suffer  a  displacement  c  from  the  pin 
center  A  of  an  amount  which  will  make  the  moment  Sc  exactly  equal  to  the  opposing 
moment  of  the  frictional  resistance  R. 


FIG.  63B. 

For  a  frictionless  pin  the  eccentricity  c  would  be  zero  and  the  stress  would  pass  through 
the  pin  center. 

Hence  from  the  figure 

Sc=Rr  and  c=r  sw.p,    .........     (63i) 

showing  that  c  is  independent  of  the  stress  S. 

When  c>e  no  rotation  will  take  place  as  a  result  of  deformation  of  the  truss  due  to 
load  effect,  and  when  c  <  e  the  member  will  turn  on  the  pin  and  the  full  amount  of  bend- 
ing cannot  be  developed  by  the  frictional  resistance. 

The  secondary  stresses  in  a  pint-connected  structure  are  found  in  precisely  the  same 
manner  as  shown  in  Art.  61,  for  riveted  connections,  except  that  for  all  members  where 


ART.  63 


SECONDARY  STRESSES 


255 


c<e  the  limiting  value  c  will  govern  and  the  secondary  stress  becomes  accordingly  less. 
Eq.  (61w)  gives  the  value  of  e=QKI/Sl,  furnishing  the  limiting  value 

^    Sic    Sir  sin  p 
=       =   ~ 


and  by  Eq.  (61v)  the  limiting  secondary  stress  becomes 

f_May_Syrsmp 
Js  —  —  j-  •       —  j  —  — 


(DDK) 


Regarding  the  values  of  p,  which  are  purely  empiric,  no  good  experimental  data 
seem  to  be  available,  though  it  is  preferable  to  choose  rather  high  values  from  12°  to  14°, 
making  sin  p  =0.2  to  0.25. 

(c)  The  effect  of  eccentric  connections.  It  frequently  happens  that  the  line  of  stress 
through  a  member  is  not  coincident  with  its  neutral  axis,  thus  giving  rise  to  eccentric 
connections. 

It  might  be  said  that  such  connections  should  never  be  tolerated,  and  hence  their 
effect  in  producing  secondary  stresses  would  not  require  investigation.  However,  there 
may  be  rare  cases  where  eccentricity  in  unavoidable  and  proper  provision  must  then  be 
made  for  taking  care  of  the  bending  stresses  thereby  produced. 

These  bending  stresses  are  not  merely  additional  stresses,  but  the  secondary  stresses 
previously  found  on  the  assumption  of  centric  connections  will  be  incorect  on  account 
of  the  changes  which  must  take  place  in  the  deflection  angles  r  and  the  angles  a,  (3,  j-, 
etc.,  between  the  truss  members,  as  a  consequence  of  the  eccentricity. 


Calling  ca  and  cb  the  eccentricities  of  the  stress  at  the  two  ends  of  a  member  AB,  Fig. 
63c,  then  the  moments  Ma=Sabca  and  M^S^  will  produce  deflection  angles  r  by 
Eqs.  (61c),  which  become 

I       t^tr  n  if  \         ^db/' 


(63L) 


256  KINETIC  THEORY  OF  ENGINEERING  STRUCTURES       CHAP,  xill 

The  values  Era  and  Erb,  resulting  from  eccentricity,  are  now  incorporated  with  their 
proper  signs  into  the  values  EJa,  and  EA$  of  Eqs.  (6lN,  o,  p)  and  the  secondary  stresses 
are'  then  computed  in  the  ordinary  way  as  described  in  Art.  61. 

Finally,  the  stresses  f8  on  the  extreme  fibers  must  be  combined  with  the  direct  stresses 
/  resulting  from  the  bending  moments  SabCa  and  SbaCb.  It  should  be  observed  that  the 
stresses  fs  and  /  frequently  have  opposite  signs,  in  which  case  the  eccentricity  effect  may 
partially  neutralize  the  secondary  stress  due  to  riveted  connections. 

It  is  sometimes  important  to  investigate  the  effect  due  to  the  eccentricity  of  the 
end  supports  where  the  friction  of  the  expansion  bearings  may  produce  considerable 
bending  stress  in  the  end  members.  Good  designs  will,  however,  obviate  any  serious 
stresses  from  this  source. 

(d)  Effect  of  temperature  and  erroneous  shop  lengths.  The  manner  of  dealing  with 
these  effects  in  the  case  of  indeterminate  structures  has  received  attention  in  previous 
chapters.  For  determinate  structures  having  end  supports  and  pin.  connections  absolutely 
frictionless,  a  small  change  in  geometric  shape,  due  to  temperature  or  erroneous  shop 
lengths,  would  have  no  appreciable  effect  on  the  stresses  in  the  members. 

However,  when  the  panel  connections  are  not  frictionless  then  any  effect,  as  unequal 
temperature,  etc.,  which  is  productive  of  a  deformation  of  the  structure,  will  also  create 
secondary  stresses. 

The  unit  stresses  /  which  result  from  any  deformation  of  the  structure  due  to  any 
changes  Al  in  the  lengths  of  the  members,  may  be  found  from  the  proportion 

Al     f     d 


whence  f=jE=Eet.-'.     .     .     ._    „     .     .'   .     .     .     (63M) 

Knowing  the  stresses  /  for  all  the  members  of  the  structure  due  to  any  simultaneous 
condition  of  temperature  or  loading,  the  values  EAa  and  K,  and  the  consequent  secondary 
stresses,  may  all  be  found  in  the  manner  described  in  Art.  61. 


ART.  64.     ADDITIONAL   STRESSES   DUE   TO   DYNAMIC   INFLUENCES 

(a)  Forces  acting  longitudinally  on  a  structure.  The  moving  loads  in  passing  over  a 
structure  exert  a  certain  horizontal  impact  which  is  transmitted  through  the  floor  system 
to  the  main  trusses.  When  the  brakes  are  set  the  amount  of  this  impact  may  become 
considerable  arid  increases  with  the  number  of  braked  wheels  on  the  span  and  .the  mag- 
nitude of  the  moving  loads.  The  forces  thus  applied  to  the  structure  are  called  tractive 
forces  and  produce  certain  additional  stresses  in  the  truss  members. 

The  effect  which  the  tractive  forces  exert  on  the  trusses  depends  on  the  relative  posi- 
tion of  the  roadway  to  the  points  of  support.  For  a  bridge,  supported  on  the  level  of  the 
loaded  chord,  all  the  additional  stresses  are  confined  to  this  chord  and  the  floor  system. 
When  the  supports  are  on  some  other  level  then  all  the  truss  members  will  be  stressed, 


ART.  64 


SECONDARY  STRESSES 


2:,7 


and  this  gives  rise  to  a  great  variety  of  cases  depending  on  the  type  of  truss  considered. 
Two  cases  will  be  presented  for  illustration. 

Let  7"=  the  tractive  force  on  one  rail  under  any  wheel  load  P. 

/=the  coefficient  of  friction  between  the  braked  wheel  and  the  rail,  usually 
taken  between  0.15  and  0.2,  and  varying  greatly  according  to  weather 
conditions. 


Then  the  total  tractive  force,  for  n  wheels  on  the  bridge,  becomes 


(64A) 


For  a  through  bridge,  the  compression  in  the  bottom  chord  members  will  then  be 
increased  by  the  tractive  force  producing  the  additional  stress  —  2  X1T,  where  x  is  the  dis- 
tance from  the  fixed  end  of  the  span  to  the  head  of  the  train. 

All  top  chord  and  web  members  remain  unaffected.  When  the  train  approaches 
from  the  roller  end  the  bottom  chord  receives  compression  and  the  stress  increases  uni- 
formly from  the  roller  end  toward  the  fixed  end. 

For  a  deck  bridge,  as  shown  in  Fig.  64A,  the  reactions  due  to  a  tractive  force  2T1 
becomes: 

A=jST;      B=-j2T;      H  =  VT (64B) 


FIG.  64A. 
The  additional  stresses  in  the  members  of  panel  1-2  are  found  as  follows: 


For  top  chord, 


17—=*- 

fi2 


h2 


IT 

V"         '*2/  ^X-*- 


I  or  bottom  chord     E*- 


M,         Ax,  -g(fto-e)-eSg!T 


cog 


For  diagonal 
For  vertical 


D  =  (A-Lsin  X)  sec  6 
F=A-Lsin  A 


(64c) 


The  angles  Jl  and  0  are  regarded  positive  as  shown  in  the  figure.  ^  The  tractive 
force  is  taken  positive  in  the  direction  from  right  to  left,  supposing  the  tram  to  appn 
the  span  from  the  roller  end  at  B. 

The  maximum  value  for  the  additional  stress  is  found  for  the  same  position  o 
train  loads  as  is  used  to  produce  maximum  primary  stress  in  any  particular  member. 


258  KINETIC  THEORY  OF  ENGINEERING  STRUCTURES       CHAP.  XIII 

These  additional  stresses  become  less  important  for  long  span,  massive,  structures 
where  the  dead  load  is  large  in  proportion  to  the  live  load.  For  smaller  structures,  the 
chord  members,  adjacent  to  the  hinged  end  of  the  span,  may  receive  excessive  additional 
stresses  requiring  increased  area  or  provision  for  reversal  of  stress. 

The  effect  of  tractive  forces  on  the  floor  system  is  not  taken  up  here. 

When  the  loaded  chord  is  on  a  grade,  the  total  weight  G  of  the  combined  dead  and 
live  loads  on  the  span  exerts  a  certain  component  G  sin  a  longitudinally  along  the  struc- 
ture and  in  a  down  grade  direction,  where  a  is  the  grade  angle.  This  component  is  trans- 
mitted to  the  hinged  support  and  should  be  considered  in  designing  this  bearing,  especially 
if  the  support  happens  to  be  a  steel  tower. 

(b)  Forces  acting  transversely  to  a  structure.  They  are  wind  pressure,  lateral 
vibrations  and  centrifugal  force  on  a  curved  track.  These  will  be  separately  discussed. 

Wind  pressure  is  the  force  due  to  wind  against  the  total  area  of  all  structural  elements 
of  both  trusses,  the  floor  system  and  live  loads,  as  presented  in  elevation.  While  the 
wind  may  have  any  direction,  it  is  presumed  that  at  times  this  direction  will  become 
normal  to  the  exposed  area  and  thus  attain  maximum  intensity. 

Experiments  have  shown  that  the  normal  wind  pressure  may  reach  a  value  of  50 
Ibs.  per  sq.ft.,  which  amount  is  assumed  in  figuring  the  wind  stresses  in  the  unloaded  struc- 
ture. 

Usually  it  would  be  unwise  to  cross  a  bridge  when  the  wind  has  sufficient  velocity 
to  produce  such  pressure,  and  hence  the  maximum  pressure  to  be  assumed  for  the  loaded 
structure  need  not  exceed  about  30  Ibs.  per  sq.ft.,  applied  to  the  total  area  of  the  struc- 
ture, the  moving  load,  and  the  portion  of  the  leeward  truss  which  is  not  covered 
by  the  moving  load.  While  both  trusses  may  not  receive  the  full  intensity  of  pressure 
it  is  customary  to  make  no  deduction  on  this  account. 

The  wind  loads  falling  on  the  truss  members  are  considered  as  concentrated  loads 
acting  on  the  several  pin  points  of  the  structure,  and  in  estimating  the  stresses  in 
the  web  members  of  the  horizontal  wind  trusses  or  lateral  trusses,  these  loads  are  treated 
as  moving  loads.  The  wind  pressure  on  the  live  load  is  always  considered  as  a  moving 
load  applied  to  the  wind  truss  of  the  loaded  chord.  The  total  maximum  wind  load  will 
govern  for  the  chord  stresses. 

The  end  reactions  of  the  wind  trusses  must  eventually  be  carried  to  the  main  truss 
supports  through  the  end  portals  or  sway  bracing.  Sometimes  only  one  wind  truss  is 
used,  and  then  the  wind  loads  of  the  other  chord  are  carried  to  this  wind  truss  by 
sway  bracing  at  each  panel  point,  throwing  the  entire  wind  load  on  the  one  lateral  system. 

Since  the  center  of  gravity  of  the  wind  load  on  the  moving  load  is  always  above  the 
plane  of  the  lateral  system  of  the  loaded  chord,  it  is  necessary  to  consider  the  overturn- 
ing effect  of  these  loads  in  producing  unequal  vertical  loading  of  the  two  main  trusses. 
Figs.  64B  show  five  cases  which  are  met  with.  In  Figs,  a,  b  and  c  one  wind  truss  is  pro- 
vided and  a  sway  brace  at  each  panel,  while  in  Figs,  d  and  e  there  are  two  wind  trusses. 

Let  Wi  =the  wind  pressure  per  panel  on  the  live  load. 

hi  =the  lever  arm  of  w\  to  the  plane  of  the  horizontal  wind  truss. 

Wz  =the  wind  pressure  per  panel  on  the  bridge. 

Ji2=the  lever  arm  of  w2  to  the  plane  of  the  horizontal  wind  truss. 


ART.  04 


SECONDARY  STRESSES 


2o9 


Then  the  overturning  moment  is 


+w2h2  in  Figs,  a  and  b  } 

fc    •    TW  i'  with  one 

—w2h2  in  lHig.  c 

in  Figs,  d  and  e  with  two  wind  trusses 


truss 


.     64n) 


and  the  vertical  load  V,  acting  down  on  the  right-hand  truss  and  up  on  the  left-hand  truss 
becomes  V=M/b.  Depending  on  the  direction  of  the  wind,  this  leads  to  positive  and 
negative  stresses  of  equal  intensity  in  each  vertical  truss.  Hence,  ±  V  is  treated  as  a 
live  load  for  the  vertical  truss  to  obtain  the  additional  stresses  due  to  overturning  effect 


rr 
h, 

i 

t—  f 

T^ 

(a) 


(c) 


(d) 


FIG.  64s. 


of  the  wind  on  the  moving  live  load,  though  it  might  be  better  to  determine  the  stresses 
separately  for  the  wind  loads  Wi  and  w2.  Calling  V\  and  V2  the  respective  vertical  wind 
loads,  pd  the  live  load  per  panel  and  gd  the  dead  load  per  panel,  then  for  parallel  chords 
the  additional  wind  stresses  become  SdVi/pd+SiV2/gd,  where  Sd  is  the  dead  load  and 
Si  the  live  load  stress  in  any  member. 

It  should  be  noted  that  the  case  in  Fig.  c  is  most  favorable,  while  the  one  in  Fig.  b 
is  most  unfavorable  in  the  production  of  loads  V,  and  may  give  rise  to  stresses  amounting 
to  as  high  as  30  per  cent  of  the  primary  stresses  in  certain  members. 

The  stresses  in  the  wind  trusses  may  be  found  by  applying  the  methods  given  in 
Chapter  XII,  and  those  occurring  in  the  portal  bracing  and  sway  system  were  analyzed 
in  Art.  62s. 

All  of  these  additional  stresses  resulting  from  wind  pressure  in  the  primary  members 
must  finally  be  combined  with  the  primary  stresses  to  obtain  total  stresses. 


260  KINETIC  THEORY  OF  ENGINEERING  STRUCTURES       CHAP.  XIII 

The  additional  wind  stresses  increase  in  importance  with  the  length  of  the  span, 
and  may  assume  enormous  proportions  in  large  structures.  It  should  be  mentioned, 
however,  that  the  frequency  in  the  occurrence  of  maximum  wind  stresses  is  extremely 
rare,  and  their  simultaneous  combination  with  maximum  primary  stresses  is  very 
improbable. 

In  view  of  these  considerations  it  would  seem  proper  to  emplo}^  higher  allowable 
unit  stresses,  say  such  as  are  permissible  for  quiescent  loading,  possibly  restricting  the 
total  sections  of  the  primary  members  more  nearly  to  those  required  only  for  maximum 
dead  and  live  loads.  For  single-track  bridges  exceeding  a  span  of  200  ft.,  and  for  double- 
track  structures  of  over  350  ft.  .span,  the  wind  stresses  must  be  fully  considered,  while 
for  smaller  structures  the  end  posts  and  end  chord  sections  should  be  appropriately  strength- 
ened, making  perhaps  little  allowance  in  the  other  members. 

Lateral  vibrations,  while  they  exist,  do  not  appear  to  produce  any  serious  increase  in 
the  primary  stresses  of  a  bridge.  This  is  certainly  true  for  highway  structures,  and  in  the 
case  of  railroad  bridges,  this  effect  may  be  reduced  to  insignificance  by  making  appropriate 
provisions  to  keep  the  moving  loads  steady  while  passing  over  the  span. 

The  magnitude  of  the  vibratory  forces  here  considered  depends  on  the  velocity  of 
the  moving  load,  the  character  and  type  of  locomotive,  the  condition  of  the  track  and 
the  relative  stiffness  and  rigidity  of  the  floor  system  and  main  trusses.  Short  spans 
and  narrow  structures,  also  skew  bridges  might  receive  considerable  vibratory  stress, 
especially  in  the  end  panels.  Coned  wheels  and  loose  or  spread  rails  are  very  objectionable. 

It  is  needless  to  say  that  factors  of  such  uncertain  constitution  and  variable  nature 
are  not  susceptible  to  analysis,  and  a  direct  measurement  (even  if  it  could  be  made)  of 
such  additional  stresses  would  merely  indicate  what  might  occur  in  certain  cases,  but 
the  results  from  different  experiments  could  not  be  accurately  correlated  with  the 
surrounding  circumstances. 

The  remedy  thus  lies  in  obviating,  so  far  as  possible,  the  conditions  producing  dele- 
terious effects  rather  than  in  any  attempt  to  estimate  their  magnitude  and  weigh  down 
the  structure  by  allowing  extra  material  in  the  required  cross-sections.  If  this  can  be 
done,  then  the  slight  additional  stresses  due  to  lateral  vibrations  may  safely  be  neglected. 
The  vertical  effect  of  impact  is  considered  elsewhere  in  this  article. 

Centrifugal  force  due  to  curved  track.  This  problem  necessarily  deals  only  with  rail- 
road bridges  and  includes  the  centrifugal  force  acting  transversely  to  the  structure  and 
producing  stresses  in  the  stringers  and  in  the  lateral  system  of  the  loaded  chord;  the 
overturning  effect  of  the  centrifugal  force  producing  stresses  in  the  stringers,  floor  beams 
and  in  the  main  trusses;  and  the  unequal  distribution  of  the  vertical  loads  on  the  floor 
and  between  the  main  trusses  due  to  the  eccentricity  of  the  curved  track  axis  with  the 
straight  bridge  axis.  This  latter  problem  was  solved  by  influence  lines  in  Art.  29,  for 
the  case  of  a  skew  plate  girder  on  a  curve,  and  will  receive  no  further  attention 
here. 

The  amount  of  the  centrifugal  force  acting  transversely  to  the  bridge  is  given 
by  the  formula 

C=^-P=cP,  (64B) 

gR 


ART.  (54 


SECONDARY  STRESSES 


261 


where  v  is  the  velocity  of  the  moving  train;  P  is  any  moving  load  on  the  bridge; 
#=32.2  ft.  per  second,  being  the  acceleration  due  to  gravity;  and  R  is  the  radius  of 
curvature  of  the  track.  The  units  are  foot,  pound,  second.  For  velocities  in  miles 
per  hour,  and  curvature  in  degrees  D,  c  =0.00001  \7v2D. 

The  following  table  gives  values  of  c=v2/gR  for  curves  of  one  to  ten  degrees  and 
velocities  which  may  be  regarded  as  maximum  for  such  curves. 


Degree  of  Curve, 
D 

v  Miles  per 
Hour. 

r 

Degree  of  Curve, 
D 

r  Miles  per 
Hour. 

c 

1 

60 

0.042 

6 

50 

0.176 

2 

58 

0.078 

7 

48 

0.189 

3 

56 

0.110 

8 

46 

0.198 

4 

54 

0.136 

9 

44 

0.204 

5 

52 

0.158 

10 

42 

0.206 

In  computing  the  additional  stresses  Sc  in  the  main  truss  members  resulting  from 
centrifugal  force,  the  maximum  total  stress  might  be  obtained  from  the  train  giving 
the  maximum  primary  stresses  rather  than  from  the  train  moving  with  the  highest 
velocity.  For  short  spans  the  maximum  combination  is  likely  to  occur  from  passenger 
trains,  while  for  long  spans  this  may  be  true  for  freight  trains  moving  at  lower 
speed. 

The  methods  of  finding  the  stresses  in  the  main  trusses  and  in  the  lateral  system 
of  the  loaded  chord  are  precisely  the  same  as  given  for  wind  loads. 

Since  the  centrifugal  force  C  is  a  linear  function  cP  of  the  moving  load,  therefore 
the  horizontal  forces  coming  on  the  lateral  truss  of  the  loaded  chord  will,  at  all  points, 
be  c  times  the  moving  load.  The  same  is  true  of  the  overturning  effect  separately 
considered,  where  the  line  of  action  of  C  is  above  the  plane  of  the  lateral  truss  and 
passes  through  the  center  of  gravity  of  the  load  P,  which 
may  be  assumed  at  the  same  level  as  the  center  of  pressure 
of  the  wind  load. 

The  stresses  in  the  lateral  truss  of  the  loaded  chord  will 
then  be  found  in  the  ordinary  way  for  loads  cP  considered 
as  moving  live  loads.  When  these  loads  P  are  uniformly 
distributed  then  the  stresses  found  for  loads  cp  will  ^be 
cp/w  times  those  found  for  the  uniform  moving  wind 
load  iv. 

The  stresses  in  the  main  trusses,  due  to  overturning 
effect  of  the  centrifugal  force  C,  are  analyzed  in  the  manner 
previously  explained  for  wind  loads.  Fig.  64c  represents  the 
condition  at  a  panel  point  assuming  the  load  P  and  the 
centrifugal  force  C  to  be  loads  per  panel  and  applied  at  the  FIG.  64c. 

center  of  gravitv  of  the  moving  load. 

The  track  is  eccentric  with  the  truss  and  the  floor  beam  is  superelevated  to  suit 


262  KIXETIC  THEORY  OF  ENGINEERING  STRUCTURES       CHAP.  XIII 

the  curved  track.  The  resultant  K,  of  P  and  C,  intersects  the  plane  of  the  lower 
laterals  at  a  distance  e  from  the  truss  axis  and  may  there  be  resolved  into  forces  P 
and  C  =cP  as  they  would  act  on  the  main  and  lateral  systems. 

The  reactions  Vi  and  V2,  representing  the  vertical  loads  which  are  transmitted 
to  the  main  trusses,  then  become 


P 
and      Vi=£ 


With  these  loads  Fj  and  V^,  which  vary  for  each  panel  point  on  account  of  the 
variable  eccentricity  e,  the  method  given  in  Art.  29  may  be  advantageously  used  to 
find  the  combined  stresses  in  the  main  trusses  by  influence  lines  and  that  is  the  best 
solution  for  this  problem.  It  should  be  noted  that  the  load  Vi  becomes  maximum 
for  quiescent  loading,  while  V%  attains  maximum  for  the  highest  velocity  of  the 
train. 

(c)  Forces  acting  vertically  on  a  structure.  This  embraces  several  causes  which 
together  exert  a  more  or  less  severe  dynamic  influence  on  the  stringers,  floor  beams 
and  vertical  trusses  of  a  bridge,  all  included  under  the  general  term  impact. 

So  far  as  these  can  be  separately  enumerated,  they  may  be  classified  as  follows: 
(1)  The  true  impact  due  to  the  instantaneous  application  of  a  moving  load  to  a  structure 
at  rest;  (2)  the  impact  resulting  from  the  unbalanced  condition  of  the  engine  drivers; 
irregularities  in  the  surfaces  of  the  rails  and  imperfections  in  the  rolling  stock,  etc.; 
(3)  elastic  vibrations  of  the  structure  which  tend  to  increase  the  effect  of  dynamic 
loads  otherwise  applied  to  a  rigid  body;  and  (4)  vertical  centrifugal  forces  occasioned 
by  the  vertical  curvature  (deflection  and  camber)  of  the  track. 

Owing  to  the  extremely  variable  conditions  and  combinations  of  circumstances  r 
it  is  practically  impossible  to  analyze  separately  the  magnitudes  of  these  several  causes 
and  their  resulting  effects,  though  a  brief  discussion  may  serve  to  point  out  precau- 
tionary measures  tending  to  reduce  the  impact  stresses  in  bridges  under  traffic.  Some 
very  valuable  tests  have  been  conducted  during  late  years,  both  in  Europe  and  America, 
which  give  a  very  good  idea  of  the  sum  total  effect  produced  on  railroad  bridges  by 
the  moving  loads,  and  from  these,  formulas  have  been  deduced  which  make  it  possible 
to  provide  suitable  strength  in  designing  bridges. 

The  additional  stresses,  resulting  from  this  combination  of  causes,  are  a  function 
of  the  velocity  and  magnitude  of  the  moving  load,  and  hence  assume  minor  importance 
when  dealing  with  highway  structures. 

The  true  impact  due  to  the  instantaneous  application  of  live  loads  is  really  not  a 
serious  source  of  additional  stresses,  because  forces  are  transmitted  through  a  structure 
with  a  rapidity  approaching  the  velocity  of  sound,  while  the  moving  train  could  not 
exert  its  effect  in  so  short  a  space  of  time. 

Theoretically,  a  load  instantaneously  applied  would  produce  a  dynamic  effect 
twice  as  great  as  the  static  load,  but  in  reality  this  cannot  take  place  and  there  will 


ART.  64  SECONDARY  STRESSES  263 

always  be  a  sufficient  lapse  of  time  between  the  application  of  the  loads  :md  tin- 
necessary  elastic  deformation  of  the  structure  to  Obviate  to  a  great  extent  this  extreme. 
dynamic  impulse.  The  vibrations  accompanying  the  moving  loads  really  constitute 
the  element  of  danger. 

The  most  unfavorable  conditions  prevail  in  short  span  structures  and  in  the  wd> 
members  subject  to  a  rapid  change  of  stress. 

The  causes  enumerated  under  (2)  and  (3)  are  really  of  the  most  serious  character, 
and  while  much  may  be  accomplished  by  a  careful  maintenance  of  the  track  and 
rolling  stock,  a  certain  average  condition  will  usually  prevail  beyond  which  no  remedial 
measures  will  be  possible  or  feasible. 

In  point  of  design  and  construction  of  a  railroad  bridge  the  following  suggestions 
may  be  offered:  The  bridge  and  its  approaches  back  for  some  distance  should  be 
straight  and  when  curves  are  absolutely  unavoidable  speed  restrictions  would  seem 
proper.  The  rails  should  be  long  and  the  joints  and  tie  connections  of  the  best  and 
most  rigid  construction,  carefully  maintained.  The  connections  at  the  abutments 
require  the  most  attention,  avoiding  uneven  settlements  and  loose  rails.  The  intro- 
duction of  continuous  roadbed  over  bridges  is  very  desirable  as  affording  some  elasticity 
to  absorb  the  impact  rather  than  to  transmit  the  same  to  the  structural  portions  of 
the  floor  and  trusses.  A  similar  effect  is  produced  by  long  ties  over  widely  spaced 
stringers.  Very  rigid  floors  and  rails  directly  on  stringers  may  be  classed  as 
objectionable. 

The  greater  the  proportion  of  dead  to  live  load  and  the  longer  the  span,  the  less 
will  be  the  effect  due  to  impact. 

The  avoidance  of  synchronous  vibrations  between  the  moving  load  and  the 
structure  is  of  the  utmost  importance  and  may  be  practically  accomplished  by  properly 
stiffening  all  the  members  and  by  providing  thorough  bracing  in  the  lateral  and  sway 
systems. 

The  effect  of  vertical  centrifugal  force  might  be  separately  estimated,  but  the 
amount  is  small  and  the  attending  conditions  are  too  uncertain  to  warrant 
this. 

As  early  as  1859,  Gerber  proposed  a  general  formula  providing  a  certain  reduction 
in  the  allowable  unit  stresses  to  cover  these  several  sources  of  dynamic  impact.  This 
formula  is 


wherein  /'  is  about  the  elastic  limit  of  the  material  and  St  and  Sd  are  the  respective 
live  and  dead  load  stresses  in  any  member. 

The  following  formula  are  at  present  in  use  and  give  the  factor  0  by  whic 
live  load  stress  should  be  multiplied,  so  as  to  include  the  dynamic  effects  considered 
under  the  present,  heading.     Calling  St  the  live   load  stress,  Sd  the  dead  load  stress 
and  I  the  strength  of  span  for  chord  members,  or  the  loaded  distance  producing 
maximum  stress  in  a  web  member,  then  <p  is.  given  as  follows: 


264 


KINETIC  THEORY  OF  ENGINEERING  STRUCTURES       CHAP,  xill 


(1)  C.  C.  Schneider,  1887,  railroad  bridges, 

,  _  1        300 


1+300      1+300 

(2)  C.  C.  Schneider,  electric  roads, 

,_t        150     _l+4oO 
1+300  ~r+300 

(3)  J.  A.  L.  Waddell,  railroad  bridges, 

,  400     _/+900 

*          Z+500  "Z+500 

(4)  J.  A.  L.  Waddell,  highway  bridges, 

,_1        100     _Z+250 
^          Z  +  150  ~r+150 

(5)  H.  S.  Pritchard,  1899,  railroad  bridges, 

,  Si         2St+Sd  t 

Y=     +  o  .  o  •    =  •  o   ,  o    ;  for  counters  0=2 

(6)  J.  Melan,  railroad  bridges, 

24 


_ 
+30       1+30 


(7)  J.  Mdan,  secondary  roads, 


(8)  F.  Engesser,  railroad  bridges, 


(9)  F.  Engesser,  highway  bridges, 

fgg  _  A  2 

^  =  l-50+y3300  •;     for     Z>33,      0  =  1.50 

(10)*  American  Railway  Eng'r  and  Maintenance  of  Way  Ass'n,  1910, 

,  _l2  +40,000 
*    12+  20,000 

Table  64A  gives  values  of  0  as  obtained  from  each  of  the  above  ten  formula  for 
a  few  even  lengths  I. 

Only  those  values  of  0  are  comparable  which  are  intended  for  the  .ame  cla«  of 
structures  as  (1),  (3),  (5),  (6),  (8)  and  (10).  It  is  clearly  seen  that  formula  (lO),ives 
the  highest  values  for  short  spans  I  and  the  lowest  values  for  long  spans,  noting  that  in 


'.     .     .     (64a) 


*  Where  I  is  the  span  length  and  all  members  receive  the  same  impact. 


ART.  64 


SECONDARY  STRESSES 


265 


this  formula  I  represents  the  length  of  span  and  not  the  loaded  distance  as  in  formula? 
(1)  to  (9). 


VALUES   OF 


TABLE  64  A. 
AS   FOUND   FROM   THE   VARIOUS  EQS.  (64c) 


I  Feet. 

(l) 

(2) 

(3) 

(4) 

(5) 

(6) 

(7) 

(8) 

(9) 

(10) 

10 

1.97 

1.48 

1.78 

1.63 

Cannot 

1.73 

1  .55 

1.97 

1.66 

1.99 

50 

1.86 

1.43 

1.73 

1.50 

be 

1.44 

1.33 

1.69 

1.50 

1.89 

100 

1.75 

1.38 

1.67 

1.40 

solved 

1.32 

1.24 

1.67 

1.50 

1.67 

200 

1.60 

1.30 

1.57 

1.28 

without 

1.24 

1.18 

1.67 

1.50 

1.33 

300 

1.50 

1.25 

1.50 

1.22 

knowing 

1.21 

1.16 

1.67 

1.50 

1.18 

400 

1.43 

1.21 

1.44 

1.18 

the 

1.20 

1.14 

1.67 

1.50 

1.11 

500 

1.38 

1.19 

1.40 

1.15 

stresses 

1.18 

1.13 

1.67 

1.50 

1.07 

600 

1.33 

1.17 

1.36 

1.13 

1.18 

1.13 

1.67 

1.50 

1.05 

l 

The  American  Railway  Engineer  and  Maintenance  of  Way  Association  formulae 
is  undoubtedly  entitled  to  the  greatest  confidence,  being  based  on  very  extensive 
experiments  which  were  carried  out  by  the  committee  on  impact  tests  under  actual 
conditions  of  traffic.  The  formula  is  not  as  yet  officially  adopted  by  the  association. 

However,  no  allowance  was  made  for  secondary  stresses  in  the  members  and  hence 
the  formula  (10)  may  be  said  to  include  more  than  actual  impact  effect  which  is 
probably  true  of  all  the  Eqs.  (64o). 

(d)  Fatigue  of  the  material.  Based  on  the  classic  experiments  of  Wohler  (1859-1870) 
which  were  continued  by  Professor  J.  Bauschinger,  a  formula  was  proposed  by  Launhardt 
(1873)  and  later  modified  by  Weyrauch,  aiming  to  apply  the  principles  of  the  fatigue 
of  material  to  the  design  of  bridge  members. 

Wohler's  law  states  that  for  a  large  number  of  repeated  load  applications,  rupture 
of  a  material  is  produced  under  stress  which  is  below  the  ordinary  (static)  breaking 
strength  of  that  material.  The  conditions  under  which  Wohler's  experiments  were 
made  differ  so  radically  from  those  attending  the  actual  operating  conditions  of  bridges, 
that  it  is  questionable  whether  anyone  is  justified  in  applying  the  Launhardt- Weyrauch 
formula?  to  bridge  designs. 

Wohler's  load  repetitions  followed  in  quick  succession  and  were  continued  without 
interruption  (several  million  times)  until  failure  was  produced.  Bridge  members  are 
subjected  to  a  repetition  of  stress  which  is  always  followed  by  a  rather  long  period  of 
rest,  and  few  structures,  even  under  the  heaviest  traffic,  are  retained  in  service  long 
enough  to  receive  several  million  applications  of  the  moving  load.  Also,  a  well-designed 
bridge  is  never  calculated  for  stresses  approaching  e ven  the  elastic  limit,  while  Wohler 
bases  all  his  observations  on  stresses  exceeding  this  limit. 

In  addition  to  these  facts,  modern  experiments  made  under  conditions  which 
respond  quite  closely  with  bridge  practice,  though  limited  in  extent,  point  uniformly 


con 


against  the  existence  of  fatigue  in  bridges. 

The  above  mentioned   Launhardt-Weyrauch  formulae  have  been  extensively  used 


266  KINETIC  THEORY  OF  ENGINEERING  STRUCTURES       CHAP.  XI I] 

in  designing  bridge  members  and  are  still  retained  in  many  specifications.  However, 
this  practice  does  not  seem  justifiable,  especially  when  allowance  is  made  for  impact  and 
secondary  stresses  generally. 

The  fatigue  formula  undoubtedly  served  a  good  purpose  in  the  days  when  so  many 
important  factors  were  neglected,  but  at  the  present  time  the  aim  should  be  to  make 
proper  allowance  for  all  the  known  stress  elements  and  thereby  reduce  the  "  factor  of 
ignorance  "  to  a  minimum. 

(e>  Unusual  load  effects  may  be  produced  by  loads  of  unusual  magnitude  applied 
at  rare  intervals  or  by  loads  which  occur  under  unusual  circumstances. 

It  is  possible  that  the  maximum  combination  of  moving  loads,  wind  pressure  and 
temperature  effects  may  take  place,  under  which  the  structure  might  be  stressed  to  its 
utmost  capacity,  while  under  ordinary  conditions  of  traffic  the  stresses  would  be  far 
below  those  for  which  the  design  was  made.  The  maximum  load  basis  for  a  design  is 
then  an  unusual  load  effect  for  the  average  use  of  a  structure. 

At  times  it  may  be  necessary  to  transport  some  abnormal  piece  of  freight  or  a  train 
of  locomotives,  which  would  exceed  the  loads  assumed  in  the  computations.  This  may 
be  done  without  danger  even  if  the  members  are  stressed  quite  near  the  elastic  limit, 
provided  the  design  is  made  with  some  forethought.  For  most  ordinary  structures  such 
a  practice  might  prove  very  disastrous. 

It  should  be  observed  that,  while  the  stress  in  any  member  for  a  particular  position 
of  a  moving  load  is  exactly  proportional  to  the  intensity  of  this  load,  the  combination  of 
maximum  and  minimum  stresses  for  such  member,  which  fixes  the  required  cross- 
section,  might  not  be  tlms  proportional.  In  the  case  of  chord  members,  all  governing 
stresses  are  proportional  to  the  loads  because  the  critical  positions  of  the  loads  are 
alike  for  all  chords,  that  is,  for  maximum  chord  stress  the  whole  span  is  fully  loaded 
while  the  minimum  chord  stress  occurs  for  dead  load  only. 

Hence,  the  chord  sections  are  directly  proportional  to  the  total  moving  load,  while 
most  of  the  web  sections  are  not  so  proportional  and  increase  more  rapidly  than  the 
loads.  This  is  particularly  the  case  with  all  counter  web  members  wherein  the  stress 
is  the  difference  between  that  produced  by  the  moving  load  and  that  due  to  the  dead 
load.  It  also  applies  to  the  sections  of  such  web  members  as  are  subject  to  stress  reversal. 

This  is  an  important  item  in  making  provision  for  a  future  increase  in  train  loads, 
where  25  to  50  per  cent  additional  carrying  capacity  might  easily  be  secured  by  providing 
a  slight  increase  in  the  counters  and  web  members  having  stress  reversals.  Many  old 
structures  would  still  be  usable  for  considerable  overload  had  such  provision  been  made 
in  their  design. 

The  best  way  to  provide  for  such  overload  in  view  of  future  increase  in  train  loads, 
is  to  design  the  structure  for  the  given  case  of  loading  and  allowable  unit  stresses  and 
then  to  increase  the  sections  of  the  counters,  and  web  members  with  stress  reversals, 
for  the  desired  overload,  which  of  course  could  not  exceed  a  reasonable  amount  according 
to  the  limitations  imposed  by  the  chord  sections. 

A  structure  so  designed  might  then  carry  say  30  per  cent  overload  without  imposing 
unduly  on  certain  few  web  members,  while  the  ordinary  structure  would  fail  by  the 
buckling  of  these  members  long  before  the  chords  had  received  any  serious  stresses. 


iART.  05  SECONDARY  STRESSES  267 

In  the  same  way  a  highway  bridge  may  be  designed  to  carry  an  occasional  overload 
W  a  certain  heavy  road  roller,  etc.,  without  increasing  the  sections  of  the  chords  and 
[main  web  members,  but  by  providing  for  the  extra  counter  stresses  due  to  the  excess  load. 

To  the  second  class  of  unusual  load  effects  belong  derailments,  collisions  with  drift- 
wood during  times  of  high  water  and  possible  settlements  of  piers  or  abutments. 

The  absolute  prevention  of  such  occurrences  may  be  classified  with  the  impossible. 
However,  the  ultimate  destruction  of  a  structure  when  they  do  occur  may  be  safeguarded 
by  applying  certain  precautionary  measures  which  should  never  be  overlooked. 

Thus  the  bridge  floor  should  be  made  sufficiently  strong  to  allow  a  derailed  train 
to  pass  over  without  breaking  through,  and  the  wheels  should  be  guided  between  guard 
rails  or  by  other  means  to  prevent  collision  with  the  main  truss  members,  necessitating 
certain  clearances  for  through  bridges  to  accomplish  this.  In  the  case  of  deck  bridges 
t  may  be  advisable  to  utilize  the  top  chord  as  a  barrier  or  protection  to  prevent  the  train 
'rom  leaving  the  structure. 

The  design  of  the  floor  should,  therefore,  aim  at  the  use  of  solid  web,  riveted 
girders  and  braces  in  preference  to  built-up  frames,  and  the  soft,  ductile  varieties  of 
steel  are  more  desirable  than  the  harder,  brittle  varieties. 

In  cases  where  high  water  may  reach  the  bottom  chord,  slender  eye-bar  members 
ire  decidedly  objectionable. 

ART.  65.     CONCLUDING   REMARKS 

(a)  Features  in  design  intended  to  diminish  secondary  and  additional  stresses.     The 

'ollowing  suggestions  should  be  given  careful  consideration: 

Skew  structures  and  bridges  on  curves  should  never  be  built  except  in  very  rare 
cases  where  no  other  disposition  is  possible.  These  types  should  be  regarded  as  measures 
)f  last  resort. 

The  axes  of  all  members  of  a  truss  should  be  in  the  same  plane  and  should  intersect 
.n  a  point  at  all  connections. 

Special  attention  should  be  directed  toward  a  careful  design  of  all  connections  with 
i  view  of  reducing  the  secondary  stresses.  Thick  gusset  plates  and  large  diameter  rivets 
materially  reduce  the  sizes  of  these  plates  and  are  thus  desirable  from  this  aspect. 

The  widths  of  members,  in  the  plane  of  the  truss,  should  not  be  chosen  larger  than 
s  absolutely  required  to  secure  proper  connections  and  stiffness  against  buckling -in 
compression  members.  It  may  thus  be  desirable  to  taper  compression  members  from 
ihe  center  toward  both  ends. 

The  cross-sections  of  members  should  be  so  chosen  that  the  material  is  concentrated 
as  far  from  the  neutral  axis  as  possible,  thus  securing  the  largest  moments  of  inertia 
for  the  smallest  over  all  dimensions.  The  cross  form  is  thus  the  least  advantageous, 
while  a  square  box  form  is  most  desirable.  In  rare  instances  the  cross  form  may  be 
acceptable,  when  the  width  of  a  member  is  determined  by  other  conditions.  Secondary 
stresses  occurring  simultaneously  in  the  plane  of  the  truss  and  in  a  cross  frame  are  additive 
in  members  of  the  box  type  while  in  the  cross  form  they  are  not  additive,  a  consideratK 
which  may  become  important  at  times. 


268  KINETIC  THEORY   OF  ENGINEERING  STRUCTURES       CHAP.  XIII 

The  secondary  stresses  may  be  reduced  to  a  certain  degree  by  shortening  all  tension 
members  and  lengthening  all  compression  members  by  amounts  Al,  determined  for  the 
case  of  maximum  total  load.  This  practice  is  generally  followed  and  provides  a  camber 
in  the  unloaded  structure  which  exactly  disappears  under  the  full  load  when  the  structure 
assumes  its  true  geometric  shape. 

The  plane  of  the  lateral  system  should  coincide  with  that  of  the  chords  and  the 
plane  of  the  floor  should  be  as  close  as  possible  to  that  of  the  lateral  system. 

The  web  members  of  the  lateral  systems  should  present  considerable  stiffness  against 
buckling,  making  l/r  not  greater  than  140  for  all  these  members,  whether  in  compression 
or  tension. 

The  end  portals,  or  cross  frames,  should  be  heavy  to  carry  the  wind  reactions  and 
dynamic  effects  to  the  supports. 

Floor  beams  should  be  made  deep  to  reduce  secondary  stresses  in  the  cross  frames- 
When  this  cannot  be  done  then  flexible  connections  should  be  provided  between  the 
truss  members  and  the  beams,  a  type  which  is  best  suited  to  large  bridges,  while  riveted 
connections  are  desirable  for  small  spans. 

The  stringers  should  be  made  heavy  and  continuous  and  should  be  designed  tot 
transmit  the  tractive  forces  to  the  panel  points  of  the  loaded  chord  instead  of  to  the  floor 
beams  by  inserting  proper  tie  members  between  the  stringers.  Long  ties  over  widely 
spaced  stringers  tend  to  relieve  impact  vibrations. 

The  use  of  pin  connections  is  not  admissible  for  bridges  of  less  than  200  ft.  span, 
and  even  in  larger  structures  the  advantages  to  be  gained  may  not  be  so  easily  demon- 
strated. Some  benefits  may  be  derived  from  pins  for  the  web  system  but  certainly  not 
for  the  compression  chord. 

According  to  prevalent  practice  in  designing  pins,  the  diameters  are  usually  so  large 
that  the  friction  produces  secondary  stresses  quite  equal  to  those  resulting  from  riveted; 
connections.  Perhaps  this  condition  might  be  remedied. 

Also  pin-connected  columns  are  not  as  stiff  as  those  with  riveted  ends,  though  the 
great  convenience  offered  by  pin  connections  during  erection  and  the  saving  of  material 
and  other  advantages  in  point  of  design,  speak  greatly  in  their  favor. 

The  advisability  of  using  the  higher  classes  of  steel  for  the  main  truss  members  of 
large  structures  and  employing  soft  steel  for  the  floor  system  was  previously  mentioned.  ' 

In  choosing  between  different  styles  of  trusses,  those  of  the  statically  determinate 
class  should  always  receive  preference,  other  things  being  equal.  The  primary  stresses 
will  usually  be  less  than  in  similar  indeterminate  systems,  especially  when  temperature 
stresses  are  included.  Yet  the  deformations  and  secondary  stresses  may  be  less  and 
frequently  the  connections  may  be  simpler  for  the  indeterminate  types. 

The  use  of  light-colored  paints  is  advisable  to  reduce  temperature  effects,  especially 
in  structures  involving  redundant  conditions. 

(b)  Final  combination  of  stresses  as  a  basis  for  designing.  If  the  several  stresses 
resulting  in  a  bridge  member  from  all  causes  could  be  absolutely  known,  then  there  is  nqj 
good  reason  why  a  design  should  not  be  based  on  a  unit  stress  of  say  four-fifths  the  elastic 
limit  of  the  material  and  still  allow  ample  leeway  for  some  deterioration  and  unusual  effects. 

The  low  unit  stresses  of  one-third  to  one-half  the  elastic  limit,  generally  employed 


ART.  65  SECONDARY  STRESSES  269 

are  intended  to  cover,  more  or  less  blindly,  the  unknown  stresses,  on  the  presumption 
that  these  are  in  a  way  proportional  to  the  known  primary  stresses.  In  reality,  few, 
if  any,  of  the  secondary  and  additional  stresses  are  directly  related  to  the  primary 
stresses,  but  are  produced  by  totally  different  causes. 

Our  knowledge  of  the  properties  of  material,  while  of  a  purely  empiric  nature, 
undoubtedly  approaches  the  truth  as  closely  as  the  knowable  accuracy  of  the  moving 
loads  would  require.  The  behavior  of  full-size  bridge  members  has  also  been  investigated 
to  an  extent  which  should  make  it  possible  to  design  such  members  with  a  reasonable 
expectancy  of  developing  a  strength  commensurate  with  that  observed  on  test  specimens. 
This  was  perhaps  not  possible  in  the  past,  but  should  be  expected  when  the  results  of  the 
elaborate  tests  of  the  American  Society  for  Testing  Materials  become  available. 

If  the  stresses  in  the  members  of  a  structure  are  determined  with  similar  exactness 
and  combined  into  totals  representing  the  maximum  and  minimum  stresses  for  each 
member  as  a  result  of  all  known  material  causes,  and  on  a  basis  of  equivalent  quiescent 
loads,  then  there  is  no  valid  argument  why  the  safe  unit  working  stress  /  should  not  be 
taken  at  least  equal  to  two-thirds  the  elastic  limit  of  the  material. 

If  also  the  main  members  and  counters,  subjected  to  reversals  of  stress,  are  designed 
for  a  30  per  cent  overload,  then  such  a  structure  should  still  be  usable  even  for  a  30 
per  cent  increase  in  the  assumed  moving  loads. 

Furthermore,  since  the  combined  effect  of  the  maximum  values  of  all  the  known 
primary,  secondary  and  additional  stresses  is  one  of  very  rare  occurrence,  this  provides 
still  greater  safety  and  subjects  the  structure  to  a  less  severe  average  usage  than  that 
intended  by  the  maximum  combination  of  all  loads. 

Assuming  then  that  the  following  stresses  have  been  computed,  or  approximately 
estimated  when  no  other  means  is  afforded,  then  the  required  area  for  any  member  may 
be  determined  from  the  formulae  given  below. 

Let  Sd  =the  stress  in  any  member  due  to  dead  load. 
Sz=max.  stress  in  this  member  due  to  moving  load. 
S'i=mm.  stress  in  this  member  due  to  moving  load. 
5,0=  the  max.  wind  stress  due  to  Wi  and  w2,  Art.  64b. 
Sr=the  stress  due  to  tractive  forces,  Art.  64a. 
St  =the  temperature  stress  when  redundant  conditions  exist. 
Sc=the  stress  due  to  centrifugal  force  from  curved  track.  Art.  64b. 
^=the  impact  coefficient  or  moving  load  factor  Eq.  (64c). 
Z=the  length  of  a  member  in  inches. 
r=the  least  radius  of  gyration  in  inches. 

/=the  allowable  unit  stress  •  =  two-thirds  the  elastic  limit  of  the  material. 
M8  =the  bending  moment  produced  by  rigidity  of  joints,  weight  of  the  member, 
eccentricity,  etc.,  representing  the  total  secondary  stress,  Arts.  61,  62,  63. 

Then  the  required  area  F  for  any  tension  member  becomes: 

t-8m4-8*+&l+-rar.  •      •      •      (65A^ 


270  KINETIC  THEORY  OF  ENGINEERING  STRUCTURES       CHAP.  XIII 

Whenever  the  area  My/fr2  can  be  added  to  the  area  given  by  the  first  term  of  Eq. 
(65A)  to  obtain  F  without  materially  altering  y  and  r,  then  it  may  be  done,  otherwise 
a  new  section  must  be  chosen  to  satisfy  the  equation.  Also,  that  combination  of  the 
stresses  in  the  parenthesis  which  gives  maximum  must  be  used. 

Eq.  (65A)  will  also  apply  to  any  compression  member  when  /  is  reduced  for  buckling 
effect  according  to  a  modified  form  of  Mr.  T.  H.  Johnson's  formula,  for  values  of  Z/r<  120: 

For  hinged  ends  and  soft  steel  of  30,000  Ibs.  per  sq.in.  elastic  limit, 

/=  20,000  -88^-  1 

T 

I  .....     •     .....     (65B) 
for  flat  ends,  /  =  20,000  -7  11-  \ 

For  hinged  ends  and  medium  steel  of  35,000  Ibs.  per  sq.in.  elastic  limit, 

/=24,000-105- 

T 

(65c) 


for  flat  ends,  /=24,000-  85- 

When  S'i  is  opposite  in  sign  from  Sd  and  Si,  then  the  member  must  be  capable  of 
carrying  an  additional  stress  l.3</>'S'i  in  the  place  of  <l>Si,  using  the  maximum  combina- 
tion of  the  same  sign  as  S'i  in  Eq.  (65A).  In  this  case,  which  is  one  of  stress  reversal, 
the  factor  1.3  should  also  be  applied  to  the  stress  </>Si.  The  impact  factors  0  and  <// 
being  determined  by  one  of  the  formula?  in  Eqs.  (64c)  for  the  load  lengths  producing 
the  respective  stresses  Si  and  S'i.  However,  this  should  not  be  construed  to  mean  that 
the  area  must  be  designed  to  carry  the  total  loads  LSflS'i+US^Si  as  if  simultaneously 
applied. 

Aside  from  the  additional  30  per  cent  in  the  live  load  stresses,  members  with  stress 
reversals  may  not  require  any  increase  in  area,  though  the  design  should  always  be  made 
so  as  to  provide  for  the  two  kinds  of  sitress. 


CHAPTER  XIV 
MITERING  LOCK  GATES 

ART.  66.     THE   NATURE    OF   THE    PROBLEM 

A  mitering  lock  gate  is  in  principle  a  three-hinged  arch,  so  placed  that  the  hinges 
have  vertical  axes,  and  the  arch  elements  are  acted  upon  normally  by  horizontal  water 
loads.  Each  half  arch  is  called  a  gate  leaf  and  the  hinges  are  replaced  by  compression 
joints  adjacent  to  the  quoin  ends  and  miter  post,  see  Figs.  6?A  and  68A. 

Each  leaf  is  designed  to  swing  about  a  vertical  axis  at  the  quoin  end,  thus  permitting 
the  gate  to  be  opened  when  the  pressures  on  up-  and  downstream  sides  are  equalized, 
a  condition  which  prevails  when  the  gate  is  swung  in  air  or  when  the  water  level  is  equal 
on  both  sides. 

The  fastening  on  the  top  of  the  quoin  post  is  called  anchorage,  and  is  usually  in 
tension  while  the  gate  is  open.  At -the  same  time  the  entire  weight  of  a  leaf  is  supported 
on  a  pintle  located  at  the  foot  of  the  quoin  post.  See  Figs.  68A. 

The  maximum  stresses  on  the  anchorage  and  the  pintle  are  encountered  when  the 
gate  is  swung  in  air,  while  the  maximum  stresses  in  the  structural  elements  of  the  gate 
occur  when  the  gate  is  closed  against  a  full  head  of  water  on  the  upstream  side  with  no 
water  on  the  downstream  side,  which  is  the  case  when  the  lock  chamber  is  empty. 

The  ordinary  working  condition  of  a  pair  of  gate  leaves  is  less  severe,  being  due  to 
a  full  head  of  water  on  the  upstream  side  and  a  counter  pressure  on  the  downstream 
side  produced  by  the  water  inside  the  lock  arid  sufficient  to  float  the  largest  vessels  for 
which  the  lock  is  intended.  These  two  water  levels  are  usually  called  the  upper  and 
lower  pool  levels. 

In  order  that  a  closed  gate  may  act  as  a  barrier  or  structural  element  against  a 
water  head,  the  bottom  horizontal  girder  is  made  to  bear  against  a  horizontal  miter  sill 
extending  over  the  full  length  of  each  leaf.  The  amount  of  pressure  exerted  by  each 
leaf  against  this  miter  sill  is  somewhat  problematic,  depending  on  the  adjustment  of 
the  sill  to  the  gate.  Usually  this  adjustment  is  so  made  that  the  pressure  is  just  sufficient 
to  make  a  water-tight  joint,  though  it  may  approach  zero  or  may  be  made  to  take  the 
entire  reaction  due  to  the  full  head  of  water. 

Hence,  a  variety  of  conditions  may  occur,  ranging  from  zero  sill  pressure  to  the  full 
hydrostatic'  head.     Usually  these  two  extreme  assumptions  are  made  and  the  resulting 
stresses  in  the  gate  are  found  for  each  case.     The  actual  condition  during  operation 
may  then  be  assumed  at  some  intermediate  state  of  sill  contact  which  can  only  1 
surmised  but  never  absolutely  known. 


272  KINETIC  THEORY  OF  ENGINEERING  STRUCTURES        CHAP.  XIV 

The  structural  elements  of  a  single  gate  leaf  consist  of  a  series  of  horizontal  girders 
connected  vertically  by  a  sheathing  or  skin  plate  and  a  number  of  vertical  intermediate 
braces  or  stiffening  girders.  The  horizontal  girders  of  both  leaves  thus  constitute  a 
series  of  three-hinged  arches  transmitting  the  water  pressure  through  the  quoin  hinges 
to  the  lock  walls.  The  vertical  system  of  stiffeners  may  be  regarded  as  a  set  of  trusses 
connected  with  the  horizontal  arches  and  thus  constituting  a  series  of  vertical  continuous 
girders,  the  supports  of  which  are  the  elastic  horizontal  girders. 

If  the  horizontal  girders  were  designed  to  carry  exactly  their  proportionate  water 
loads  and  at  the  same  time  be  allowed  to  deflect  in  proportion  to  these  loads,  then  for 
no  contact  between  the  miter  sill  and  the  vertical  stiffeners,  the  entire  work  would  fall 
on  the  horizontal  girders  and  the  vertical  stiffeners  would  do  no  work  and  would  thus 
be  superfluous. 

However,  for  practical  reasons,  the  top  horizontal  girders  are  always  designed  much 
stronger  than  is  actually  required  for  the  water  load  and  also  the  other  girders  are  not 
all  different,  owing  to  commercial  sizes  of  rolled  shapes  and  the  saving  in  manufacture 
resulting  from  duplication  of  like  parts.  Also,  some  sill  pressure  must  be  provided  to 
accomplish  water  tightness  of  the  gate.  Therefore,  the  vertical  system  of  stiffeners 
must  necessarily  do  a  certain  amount  of  work  in  distributing  these  inequalities  between 
the  horizontal  girders  and  the  vertical  stiffening  system.  The  reactions,  which  the 
horizontal  girders  thus  present  to  each  vertical  girder,  may  be  considered  as  redundant 
reactions  of  a  vertical  continuous  girder  on  n  flexible  horizontal  supports,  where  n  is 
the  number  of  horizontal  arch  girders. 

The  vertical  girders  are  not  very  definite  structural  elements,  being  made  up  of  a 
rather  disconnected  system  of  web  bracing  between  the  horizontal  arches.  The  total 
vertical  stiffness  of  a  gate  is  thus  produced  by  the  combined  effect  of  the  sheathing 
plates,  the  quoin  and  miter  posts  and  the  intermediate  vertical  girders.  If  these  were 
all  combined  in  one  vertical  system  so  as  to  represent  one  vertical  girder  for  a < gate  leaf, 
then  the  average  stiffener  per  foot  of  gate  could  be  approximated  by  dealing  with  an 
average  vertical  section  of  the  gate  leaf.  Such  a  hypothetical  vertical  girder  could  then 
be  treated  as  a  continuous  girder,  on  n  horizontal  supports  consisting  of  n  three-hinged 
arches. 

The  problem  would  thus  be  to  find  the  magnitudes  of  these  n  redundant  supports 
Xi  to  Xn  of  the  vertical  girder  for  a  certain  water  load  (usually  the  maximum)  together 
with  the  elastic  displacements  d\  to  dn  of  the  n  supports. 

Since  the  contact  at  the  miter  sill  cannot  be  absolutely  adjusted,  it  is  best  to  inves- 
tigate two  cases;  one  with  full  contact  against  the  miter  sill,  giving  the  maximum  sill 
pressure;  and  the  other  where  the  miter  sill  pressure  is  just  zero  and  there  is  theoretically 
no  contact. 

It  is  certain  that  all  possible  cases  must  be  included  between  these  two  extremes. 
However,  in  practice  it  is  customary  to  design  the  bottom  horizontal  girder  to  carry 
the  total  water  load  without  sill  contact  and  to  design  the  sill  amply  strong  to  carry  the 
same  full  load,  but  to  adjust  the  gate  so  that  the  sill  pressure  is  only  sufficient  to  make 
a  water-tight  joint. 

The   author  has  proposed   details   of   construction   obviating   such   duplication   of 


ART.  67  MITERING  LOCK  GATES 


273 


strength  in  the  gate  and  miter  sill  by  the  introduction  of  a  flexible  sill  connection  whereby 
the  water  pressure  against  the  sill  can  never  exceed  a  certain  allowable  quantity. 

ART.   67.     THE   THEORY  OF  THE  ANALYSIS 

The  principal  system.  The  problem  is  analyzed  according  to  the  solution  of  the 
general  case  of  redundancy  presented  in  Art.  55,  using  the  graphic  method  for  all  deflection 
polygons.  Since  there  are  only  two  cases  of  loading  requiring  investigation  it  is  best  to 
employ  the  general  Eqs.  (55A)  and  (55B)  necessitating  two  solutions,  each  involving  n 
equations  of  n  unknown  quantities  X. 

The  effect  of  temperature  will  be  neglected  in  the  present,  though  it  could  be 
determined  in  accordance  with  Art.  56. 

To  make  the  problem  perfectly  definite,  it  is  first  necessary  to  decide  on  the  principal 
system  with  its  determinate  reactions,  and  then  to  apply  the  redundant  forces  along  with 
the  external  loads  following  the  general  scheme  illustrated  in  Figs.  7  A  to  ?E,  Art.  7. 

The  case  of  full  contact  at  miter  sill  is  illustrated  in  Figs.  67A  and  B. 

Fig.  67A  shows  a  lock  gate  in  plan,  with  the  water  loads  p,  per  linear  foot,  applied 
to  the  upstream  surface.  The  resultants  R,  of  the  loads  p  on  each  leaf,  give  rise  to 
arch  thrusts  R'  and  R"  at  the  quoin  ends  and  a  horizontal  thrust  H  acting  between  the 
miter  posts.  The  gate  is  supposed  to  undergo  elastic  deformations,  as  indicated  by  the 
dotted  deflection  curves.  The  lock  walls  take  up  the  arch  thrusts. 

Each  horizontal  girder  receives  a  total  load  P  which  is  proportional  to  the  depth 
of  water  below  the  surface.  These  loads  are  easily  computed  for  all  the  horizontal  girders 
and  are  assumed  to  be  known. 

Fig.  67fl  shows  a  projected  elevation  of  the  gate  with  the  several  loads  P  applied 
to  the  respective  horizontal  girders.  The  bottom  girder  rests  against  the  miter  sill  over 
its  entire  length. 

The  vertical  girder  AB  represents  an  element  one  foot  thick  lengthwise  of  the  gate 
leaf  and  may  be  regarded  as  a  plate  girder  whose  chords  are  portions  of  the  sheathing 
plate  and  all  material  making  up  the  average  vertical  stiffness  of  the  leaf,  while  its  web 
may  be  neglected. 

This  vertical  girder  is  now  treated  as  the  principal  system  and  the  problem  consists 
of  determining  the  stresses  in,  and  the  deflection  of,  this  girder. 

The  determinate  supports  for  this  vertical  girder  (or  principal  system)  are  the  sill 
reaction  B,  which  is  immovable  in  the  present  case,  and  the  reaction  A  offered  by  the 
top  horizontal  arch  girder.  All  the  other  horizontal  girders  2  to  n  (shown  by  dotted 
lines)  are  now  considered  removed  and  the  redundant  reactions  X  which  they  present 
to  the  vertical  girder  are  shown  as  external  forces  Xi  to  Xn. 

The  principal  system  is  thus  the  vertical  giider  on  two  determinate  supports  A 
and  B,  with  A  elastic  and  B  immovable.  The  external  loads  on  the  principal  system 
are  the  water  loads  P  and  the  redundant  reactions  X. 

The  yielding  condition  of  the  support  A  may  be  obviated  by  introducing  another 
redundant  reaction  Xi  and  including  the  top  horizontal  arch  as  a  part  of  the  principal 
system  with  immovable  supports  at  the  quoin  ends  D  and  E.  The  reaction  A  is  then 


274 


KINETIC  THEORY  OF  ENGINEERING  STRUCTURES        CHAP.  XIV 


-^^-::'^' 

"                     ] 

T 

'I                     J 

ii        L°CI 

1  | 

1 

A\ 

i 

1 

! 

FIGS.  67A,  B — Principal  System,  Full  Loading,  for  Case  of  Full  Contact  at  Miter  Sill. 


<-•'_£" 

&** 


FIG.  67c— Principal  System,  Full  Loading,  for  Case  without  Contact  at  Miter  Sill. 


AET.  67 


MITERING  LOCK  GATES 


275 


regarded  as  the  resultant  of  the  immovable  reactions  A'  and  A",  and  the  deflection  di  of 
the  point  of  application  of  Xlf  can  then  be  evaluated  in  terms  of  Xit  making  3l  =di  ^X^ 
The  deflection  curve  of  the  vertical  girder  is  indicated  by  a  heavy  dotted  line. 


FIG.  67o — Principal  System,  Condition  X=0.     For  Case  of  Full  Contact  at  Miter  Sill. 

The  case  without  contact  at  the  miter  sill  is  illustrated  in  Fig.  67c,  where  the  principal 
system  consists  of  the  vertical  girder  A  B  supported  on  two  yielding  supports.  Herwe, 
by  taking  in  the  top  and  bottom  horizontal  arches  as  part  of  the  principal  system  the 
fixed  reactions  become  the  four  points  D,  E,  F  and  G,  while  the  elastic  supports  A  and 


FIG.  67E-Principal  System,  Condition  X,=  l.     For  Case  of  Full  Contact  at  Miter  Sill. 

B  undergo  certain  deflections  ^  and  90,  which  are  evaluated  in  terms  of  the  redundants 
A'i  and  X6      The  reaction  B  is  the  resultant  of  the  arch  thrusts  B'  and  B  . 

This  case  presents  no  new  difficulty  except  to   add  one  redundant    condition   Xn 


276 


KINETIC  THEORY  OF  ENGINEERING  STRUCTURES       CHAP.  XIV 


and  another  water  load  Pn  to  those  involved  in  the  previous  case.  No  further  con- 
sideration is,  therefore,  given  to  the  case  without  sill  contact. 

The  conventional  loadings  of  the  principal  system  are  illustrated  in  Figs.  67o  and  E, 
for  the  case  of  full  contact  at  the  miter  sill. 

In  Fig.  6?D  the  condition  X  =0  is  represented.  This  embraces  the  total  water 
loads  on  the  principal  system  and  the  determinate  reactions  A0  and  B0  resulting  there- 
from, when  all  the  redundants  X  are  removed. 

Fig.  6?E  shows,  for  example,  the  condition  X3  =  l,  where  all  loads  are  removed  from 
the  principal  system  except  a  unit  load  applied  in  the  place  of  ^3  and  acting  in  a  direction 
opposite  to  the  direction  assumed  for  the  redundant  reaction  X3.  Similarly  all  other 
conventional  loadings  from  Xi  =1  to  Xn  =  \  are  applied  one  at  a  time.  It  was  not  deemed 
necessary  to  show  them  all  by  figures. 

The  external  forces.  Eqs.  (55A)  may  now  be  applied  to  the  principal  system  as 
above  illustrated,  choosing  the  case  of  n  horizontal  arches  with,  or  without,  miter  sill 
contact.  The  only  distinction  between  the  two  cases  is  that  Xn=0  for  full  contact, 
otherwise  the  general  formula  apply  to  both  cases. 


r 


* 

p, 

5 

5 

* 

6 

*,„ 

p 

P 

I 

x, 

i 

3 

• 

4 

x. 

'       5 

m 

. 

m* 

, 

1 

•  

X 

^rv 

n 
X 

V 

"m 

1           » 

u 

v 

"3 

V,      . 

2 

B 


FIG.  67F. 


Let  Fig.  67r  represent  a  general  case  of  n  redundants,  showing  the  principal  system 
on  two  determinate  supports  A  and  B.  Also,  let  m  be  any  panel  point  about  wrhich 
the  moment  Mm  may  be  desired  to  determine  the  stresses  in  the  chords  of  the  vertical 
girder. 

Eqs.  (55A)  for  n  redundants  and  moments  about  m  become: 


A=A0  -AlXl  -A2X2  -A3X3 
B=B0  —B\X\  —B2X2  —B^Xs 
m  =  Mmo-MmlX!  -Mm2X2-Mm3X3 


•  •   — AmXm     .  .   — AnXn 

•  •   —BmXm      .  .   —BnXn 


(67A) 


The  lettered  constants  in  Eqs.  (67A)  are  evaluated  as  follows: 


ART.  67 


MITERING  LOCK  GATES 


277 


Condition. 


Xm=l 


A  --— 

_!'A* 

AS=  /*7 

Am  =  ~h~ 

1-hm+i 

Am+i  =  — ^ 

A»  =  0 


=  0 


/  ^2\ 

'.  =  hm  I  1  —  7"  ) 

V      hS 


--^(l-h.) 


M 


The  tabular  values  substituted  into  Eqs.  (67 A)  give  after  considerable  reduction: 


.  .    .    .     (67B) 


The  last  of  Eqs.  (67s)  for  m  =  l  and  hm=hlf  gives  MA=0;  and  for  m=n  and 
hm=hn  =0,  gives  Mfi=0,  which  results  must  follow  from  the  conditions  of  equilibrium. 

The  same  expression  for  Mm  may  also  be  obtained  by  taking  the  moments  about 
m  of  all  the  external  forces  to  either  side  of  a  section  through  m. 

The  redundant  reactions  X.  The  general  work  Eqs.  (55n),  when  the  effect  of  tempera- 
ture is  neglected,  will  furnish  n  equations  for  finding  n  redundant  reactions  X,  as  follows: 


etc., 


etc.,  etc., 

+X2dn?.    +X33n3 


2_4+etC.,+^"n^2n 

etc.,  etc.,         etc.,  to 


(67c) 


wherein  the  subscript  m  has  all  values  successively  from  1  to  »      In  all  ^es  of  double 
subscripts  the  first  one  always  indicates  the  point  of    application  of  the  conventional 


278  KINETIC  THEORY  OF  ENGINEERING  STRUCTURES        CHAP.  XIV 

load  while  the  second  one  refers  to  the  location  of  the  deflection.  It  should  also  be 
remembered  that  all  deflections  d,  with  like  double  subscripts,  are  always  equal;  thus 
^2-6=^6-2  and  dm5=§5m,  etc. 

The  double  subscript  bearing  d's  are  found  from  deflection  polygons  drawn  for  the 
principal  system,  and  for  each  case  of  conventional  loading,  requiring  n  such  deflection 
polygons  for  n  redundants.  Thus,  the  several  double-subscript-bearing  deflection  ordinates 
3  of  the  first  of  Eqs.  (67c)  are  all  obtained  from  a  deflection  polygon  drawn  for  condition 
Xi  =1  acting  on  the  principal  system. 

The  deflections  8l}  d2,  <53  to  3n  are  the  actual  displacements  of  the  points  1  to  n,  of  the 
principal  system.  Since  these  points  are  the  points  of  support  offered  by  the  horizontal 
arch  girders  to  the  vertical  girder,  they  represent  the  deflections  really  produced  in  the 
arch  girders  by  the  redundant  reactions  ,X\  to  Xn  respectively.  Hence  di,  32,  £3  to  dn 
cannot  be  deternined  until  the  redundants  Xi  to  Xn  are  known,  though  they  may  be 
expressed  in  terms  of  the  latter  without  increasing  the  number  of  unknowns. 

Let  di  =the  deflection  of  horizontal  girder  I  at  the  point  of  support  of  A"i  and  due 
to  the  conventional  load  X\=l. 

d-2.  =the  deflection  of  horizontal  girder  II  at  the  point  of  support  of  X%  and  due  to  the 
conventional  load  ^2  =  1.  Similarly  for  ^3  to  dn.  When  several  horizontal  girders  are 
of  like  sections  then  their  deflections  d  become  equal,  thus  obviating  the  necessity  for 
drawing  n  such  deflection  polygons. 

When  the  conventional  deflections  d\,  d%,  d%  to  dn  are  known,  then  the  actual  deflec- 
tions di,  82,  §3  to  dn,  due  to  the  redundant  reactions  Xi  to  Xn,  become  (by  the  law  of  pro- 
portionality between  cause  and  effect) : 

fli-di^i;  82=d2X2;  83=d3X3;  etc.,  to  8n=dnXn.       .     .     .'     (6?D) 

It  should  be  noted  that  in  the  particular  case  here  considered,  the  deflection  d\  =d\-\ 
because  the  vertical  girder  is  not  deflected  by  the  load  X\=\  while  the  total  displacement 
of  the  principal  system  at  point  1  is  that  due  to  the  deflection  of  the  horizontal  arch,  and 
the  vertical  girder  merely  moves  with  the  .arch.  Similarly  when  there  is  no  sill  contact 
then  dn=8nn.  These  conditions  make  it  possible  to  draw  the  deflection  polygons  for 
X\  =  1  and  Xn  =  1  for  the  vertical  girder. 

The  conventional  loading  for  the  principal  system  (being  the  vertical  girder  of  1 
ft.  thickness  measured  lengthwise  of  the  horizontal  girder)  is  taken  as  a  unit  force  of  say 
one  kip  =  1000  Ibs.  The  corresponding  conventional  loading  for  the  horizontal  arch 
girders  should  then  be  one  kip  per  foot  of  wetted  length  of  the  arch.  Whatever  loads  are 
used,  this  ratio  between  the  conventional  loads  of  the  horizontal  and  vertical  systems 
must  be  maintained. 

The  values  from  Eqs.  (6?D)  are  substituted  into  Eqs.  (67c),  which  latter  are  then 
solved  by  any  process  of  elimination  to  obtain  the  n  redundants  X. 

When  these  redundants  X  are  thus  evaluated  for  any  particular  case  of  water  loads 
PI  to  Pn,  then  the  functions  A,  B,  Mm  and  di  to  dn  may  all  be  obtained  by  substitutions 
in  Eqs.  (67B)  and  (67o) ,  and  the  problem  is  considered  solved. 


ART.  08 


MITERING  LOCK  GATES 


279 


The  moment  Mm,  will  in  turn  furnish  the  chord  stress  in  the  vertical  girder  for  the 
center  of  moments  m.  The  reaction  B  will  represent  the  pressure  against  the  miter  sill 
which  must  be  considered  in  designing  this  sill. 


ART  68.     EXAMPLE-UPPER   GATE,   ERIE    CANAL   LOCK   NO. 


22 


The  example  originally  chosen  for  analysis  was  one  of  the  huge  Panama  Canal  lock 
gates  with  sixteen  horizontal  arches.  After  completing  the  drawings  and  computations 
it  was  deemed  advisable  to  use  a  smaller  gate  for  illustrative  purposes  as  the  theory  would 
be  less  obscured  by  the  rather  extensive  tabulated  computations.  Accordingly  one  of 
the  small  1909  steel  gates  of  the  Erie  Canal  was  used. 

Figs.  68A  show  one  leaf  of  this  gate  in  elevation  and  two  sections.  There  are  six 
horizontal  girders  and  the  upstream  side  is  covered  by  a  f-inch  sheathing  plate,  while  the 
downstream  side  is  open. 

The  material  is  soft  steel  with  a  modulus  of  elasticity  #=29,000  kips  per  sq.in.  The 
quoin  and  miter-post  cushions  are  made  of  white  oak  which  when  saturated  with  water 
becomes  quite  soft.  According  to  the  best  available  data,  the  modulus  EI,  for  such 
material,  may  be  taken  at  about  30  kips  per  sq.in. 

The  cross-sections  of  the  horizontal  girders  are  shown  in  Figs.  68B,  from  which  it  is 
seen  that  girders  II  to  V  are  exactly  alike,  so  that  there  are  only  two  different  types  of 
horizontal  girders  requiring  analysis. 

Since  the  sheathing  plate  is  continuous  between  the  girders,  a  certain  portion  of  this 
plate  will  act  with  the  flange  material  of  each  horizontal  girder.  It  is  difficult  to  say 
just  how  much  of  the  sheathing  will  actually  act  in  that  manner,  but  in  the  present  analysis 
a  width  of  sixteen  inches  is  thus  included  with  the  upstream  flanges  of  the  horizontal 
girders  except  for  girder  I,  where  the  sheathing  stops  at  the  girder  web,  and  only  eight 
inches  are  included  here. 

The  cross-sections  F  and  moments  of  inertia  I,  of  these  horizontal  girder  sections, 
are  computed  from  Figs.  68B  and  given  in  Table  68A. 

TABLE  68A 
VALUES  OF  F  AND  I  FOR  THE  HORIZONTAL  GIRDERS. 


At  Points 

Girder  I. 

Girders  II  to  V. 

F 
sq.in. 

7 
in.1 

F 

sq.in. 

/ 

in.< 

2  and  10 

3  to  9 

26.45 
26.91 

3740.6 
4120.1 

30.96 
31.30 

4054.6 
4466.0 

Fig.  68c  shows  the  average  vertical  girder  and  the  water  loads  for  the  gate. 
This  girder  is  made  up  of  the  material  which  gives  vertical  stiffness  to  the  gate  and 
includes  the  sheathing  plate  on  the  upstream  surface  and  such  plates  and  angles  as 


280 


KINETIC  THEORY  OF  ENGINEERING  STRUCTURES        CHAP.  XIV 


GENERAL  DRAWING  OF   MITERING  LOCK  GATE 


DOWNSTREAM  ELEVATION. 


wood 

Cushio 


DOWNSTREAM  SIDE  OPEN. 

SECTION  ON  GIRDER  IV. 


FIG.  68A. 

SECTIONS   OF  HORIZONTAL  GIRDERS. 
GIRDER   I.  ,  GIRDERS   IITO,V. 


—  Z9"       •• 

POINTS  ZfclO. 


AT  POINTS  3T09. 

FIGS.  68s. 


29 

AT  POINTS  2  It  io. 


AT  POINT*  3  TO  9. 


ART 


MITERING  LOCK  GATES 


281 


extend  vertically  over  the  downstream  surface.     The  total  web  mav  be  considered 

^  f  the  leaf 


The  total  volume  of  this  material  is  then  divided  by  the  length  of  the  gate  leaf 
(24.3  ft.)  giving  the  average  section  of  the  vertical  girder  per  horizontal  foot  of  the  gate 


r-roo.5 


i  -30.3— * 
ATCRSURTAC     _ 


MITER  SILL. 


The  loads  p  are  th«  water  pressures  in  kips  per  foot- of  horizontal  arch: 
The  loads  P=  27  parrthe  total  pressures  for  the  horirontal  girders. 

FIG.  68c. 

is  shown  in  Fig.  68c.     The  vertical  girder  is  thus  considered  as  an  average  strip  of  the 
;ate,  one  foot  thick  lengthwise  of  the  leaf. 

The  water  loads  p  are  the  actual  pressures  (in  kips)  per  horizontal  foot  of  gate  and 
xtend  vertically  from  center  to  center  of  the  panels  between  the  arches.  The  loads 
*=-27p  are  the  total  pressures  for  the  horizontal  girders  of  wetted  length  27  ft.  Hence 
ili-P  would  be  the  total  water  pressure  on  one  gate  leaf,  while  £?p  would  be  the  total 
3ad  on  the  vertical  girder  with  an  exposed  surface  of  1  •  ho. 


282 


KINETIC  THEORY  OF  ENGINEERING  STRUCTURES        CHAP.  XIV 


The  deflections  d  of  the  horizontal  arches  are  now  found  for  the  conventional  loading 
according  to  the  previous  article.  This  loading  should  be  one  kip  per  horizontal  foot 
of  wetted  length  of  the  arch,  corresponding  to  a  conventional  loading  of  one  kip  for  the 
vertical  girder  one  foot  wide  horizontally.  However,  in  the  present  problem  a  conven- 
tional load  of  q  =  l/l  -=1/27  =0.037  kip  will  be  used  for  the  horizontal  girders,  giving 
deflections  l/l  times  too  small.  That  is,  the  conventional  load  should  have  been  I  kips 
on  each  leaf,  uniformly  distributed,  but  instead  of  this  one  kip  was  distributed  over  each 
leaf  of  27  feet  wetted  length. 

The  deflection  of  a  three-hinged  arch  under  a  uniformly  distributed  load  is  made 
up  of  a  direct  cross  bending  of  the  beam  or  half  arch,  and  a  deflection  of  the  arch  crown 
due  to  axial  compression,  as  shown  in  Fig.  68D. 

Let  Nav  be  the  average  axial  thrust  in  the  half  arch  of  length  I  and  cross  section  F. 
Also,  let  dc  be  the  deflection  at  the  crown  normally  to  DC  due  to  a  shortening  Jl  in  the 
length  /  as  a  result  of  the  compression  Nav. 

Then 


i          *  V  fl  ft  j 

=  EF       and 


__ 

tan  a     EF  tan  a ' 


(68A) 


FIG.  680. 

The  effect  of  dc  on  the  arch  is  merely  to  lower  the  crown  to  C' ,  leaving  the  beams 
DC  and  CE  otherwise  unchanged  and  producing  the  dotted  deflection  line  DC'E.  With 
this  must  be  combined  the  deflection  due  to  bending  and  shear  in  the  beam,  finally 
producing  the  deflection  curve  DC'E,  with  ordinates  d  measured  normally  to  the  girder 
DC.  The  bending  deflection  will  be  determined  from  the  bending  moments,  by  the 
method  given  in  Art.  39  for  solid  web  beams. 

Before  proceeding  to  this  it  may  be  well  to  say  a  few  words  regarding  the  particular 
deflection  d  which  is  to  be  used  in  the  solution  of  the  problem,  since  this  point  may 
involve  a  considerable  element  of  doubt. 

In  the  first  place  the  deflections  must  be  determined  in  the  direction  normally  to 
the  girder  DC,  because  that  is  the  direction  of  action  of  the  water  loads  P  and  of  the 
redundant  reactions  X.  Ordinarily  the  deflections  for  any  three-hinged  arch  would  be 
taken  normally  to  the  span  DE. 

The  principal  system  was  chosen  as  an  element  of  average  vertical  cross-section 
of  the  gate  and  the  deflection  properly  belonging  to  such  a  hypothetical  vertical  girder 


ART.  08 


MITERING  LOCK  GATES 


283 


would  necessarily  be  an  average  deflection.  Hence,  the  value  of  d  which  is  actually 
used  is  obtained  by  dividing  the  area  DCC',  of  the  deflection  curve  DC*,  by  the  length 
/  of  the  leaf.  This  average  deflection  dav,  is  shown  in  Fig.  68D  under  the  assumed  vertical 
girder.  See  also  Figs.  68r. 

The  bending  moments  for  the  various  points  of  a  horizontal  girder,  due  to  the  load 
9 -=0.037  kip  per  foot  of  wetted  length,  are  now  determined  by  drawing  a  resultant 
polygon  through  the  girder  and  then  taking  the  moments  of  the  resultant  normal  thrust 
about  the  center  of  gravity  of  the  section,  as  shown  in  Fig.  68E. 

The  loads  </i  to  q<n  are  first  combined  into  a  force  polygon  furnishing  the  resultant 
R  and  the  reactions  R'  and  H,  which  latter  can  be  determined  when  the  points  D  and  C 
through  which  the  resultant  polygon  shall  pass,  are  fixed.  The  point  C  is  taken  in  the 
center  of  the  joint  and  the  point  D  is  taken  so  that  the  resultant  R'  is  normal  to  the 
hollow  quoin.  The  resultant  polygon  is 'then  easily  drawn  and  the  offsets  v,  between 
it  and  the  gravity  axis  of  the  girder,  are  thus  found.  The  axial  thrusts  N  are  then  scaled 
from  the  force  polygon  and  from  these  and  the  eccentricities  v  the  moments  M  =Nv  are 
computed  in  Table  68s . 

The  elastic  loads  w=MAx/l  are  also  determined  and  from  these  the  required 
deflection  polygon  is  drawn. 

TABLE  68s 
ELASTIC  LOADS  w  FOR  HORIZONTAL  GIRDERS. 


Pt. 

TV 
Kips. 

• 
inches. 

M  =  Nv 
Kip  in. 

Horizontal  Girder  I. 

Horizontal  Girders  II  to  V. 

/ 
in.4 

M 

I 

Ax 
Inches 

M4x 

"•"7" 

/ 
in.< 

M 
I 

4x 
Inches. 

Mix 
„__ 

D,  1 

1.38 

0.0 

0.00 

0.0 

0.0 

40.00 

0.0126 

40.00 

0.0116 

2 

1.396 

1.7 

2.373 

3740.6 

0.00063 

24.00 

0.0377 

4054.6 

0.00058 

24.00 

0.0348 

3 

1.40 

7.4 

10.360 

4120.1 

0.00251 

32.64 

0.1142 

4466.0 

0.00232 

32.64 

0.1054 

4 

1.40 

13.2 

18.480 

4120.1 

0.00449 

32.64 

0.1643 

4466.0 

0.00414 

32.64 

0.1514 

5 

1.40 

16.4 

22.960 

4120.1 

0.00558 

32.64 

0.1877 

4466.0 

0.00514 

32.64 

0.1728 

6 

1.40 

17.4 

24.360 

4120.1 

0.00591 

32.64 

0.1845 

4466.0 

0.00545 

32.64 

0.1697 

7 

1.40 

15.8 

22  .  120 

4120.1 

0.00537 

32.64 

0.1542 

4466.0 

0.00495 

32.64 

0.1421 

8 

1.40 

12.0 

16.800 

4120.1 

0.00408 

32.64 

0  1000 

4466.0 

0.00376 

32.64 

0.0920 

9 

1.40 

6.0 

8.394 

4120.1 

0.00204 

24.00 

0.0266 

4466.0 

0.00188 

24.00 

0.0246 

10 

1.39o 

0.5 

0.698 

3740.6 

0.00018 

33.00 

0.0030 

4054.6 

0.00017 
0  0 

33.00 

0.0028 

C,  11 

1.37o 

0.0 

0.0 

0.0 

316.84 

0.9848 

316.84 

0.9072 

284 


KINETIC  THEORY  OF  ENGINEERING  STRUCTURES       CHAP.  XIV 


0.5 


I  KIR 


FIG.  68E. 


ART.  68  MITERING  LOCK  GATES  285 

There  are  two  different  kinds  of  horizontal  girders,  thus  making  two  such  deflection 
polygons  decessary,  one  for  Girder  I  and  one  for  Girders  II  to  V.  The  sections  are  given 
in  Figs.  68s  and  Table  68A.  The  increments  Ax  are  measured  along  the  span  Z)C  such 
that  2Ja;=L  All  dimensions  are  now  taken  in  inches  and  loads  in  kips. 

According  to  Art.  39,  an  equilibrium  polygon  drawn  for  the  elastic  loads  w  with  a 
pole  distance  equal  to  £"=29,000  kips,  will  represent  the  deflection  polygon  for  the 
moments  M,  with  ordinates  measured  to  the  scale  of  lengths  of  the  drawing. 

In  Figs.  68r,  the  pole  distances  were  made  equal  to  H=  E/20  ,000  =  1.4  5,  giving 
deflections  20,000  times  actual  to  the  scale  of  lengths. 

The  moment  deflection  polygon  is  constructed  by  plotting  the  values  M/I,  from 
Table  68s,  as  ordinates  at  the  points  1  to  11,  using  any  convenient  scale,  then  finding 
the  centers  of  gravity  g  of  the  several  trapezoidal  areas  of  this  M/I  polygon  and  applying 
the  loads  Wi  to  WIQ  at  these  centers.  The  equilibrium  polygon  drawn  for  the  w  forces 
is  then  the  moment  deflection  polygon,  with  deflections  measured  parallel  to  the  direction 
taken  for  the  w  loads,  in  this  case  perpendicular  to  the  chord  of  the  girder. 

The  scale  for  the  loads  and  the  force  polygon  is  immaterial  so  long  as  it  is  convenient 
and  the  pole  distance  H  must  be  laid  off  to  the  scale  of  loads.  The  pole  0  may  be 
anywhere  to  suit  a  convenient  closing  line  D'C'. 

The  actual  moment  deflection  of  girder  I  at  point  6  is  thus 

-0.00188  ind, 

The  actual  deflection  due  to  shear  at  the  same  point  6,  which  is  the  point  of  maximum 
moment,  is  by  Eq.  (39D)  , 

5X24'36 


= 
6 


_ 
2EFi     2X29000X30X0.370 


=0.00019  inch, 


where  Fj  is  the  area  of  the  web  plate  and  M6  is  taken  from  Table  68s. 

The  deflection  for  point  6,  due  to  shear  alone,  is  thus  about  10  per  cent  of  the 
deflection  for  the  same  point  and  due  to  moments.  Hence,  all  the  deflection  ordinates 
are  increased  10  per  cent  to  obtain  the  combined  moment  and  shear  deflection  polygon. 

The  crown  deflection  dc,  Eq.  (68A),  due   to  rib  shortening  by  axial  thrust,  is  now 

found.  .       . 

Since  a  portion  of  the  gate,  at  the  quoin  and  miter  posts,  is  made  of  oak,  which  i: 
probably  saturated  when  the  gate  is  in  operation,  it  is  necessary  to  make  allowance  for 
this  condition  in  computing  Al. 

For  an  average  thrust  JV-  1.4  kips,  and  a  net  Z-=  292  inches,  F  -26.91  sq.m.,  and 
£=29  000  kips  per  sq.in.  for  the  steel  part  of  the  gate  and  Z'=24.8  inches,  F'=24 
=864  sq.m-,    and    El  =30  kips    per    sq.in.    for  saturated  oak,  then   for  tan  a  =0.367. 
Eq.  (68  A)  gives  for  the  steel  and  oak  portions 

1.4  X292  1.4X24.8       =0.00499  inch, 

de  =  29000  X26.91  X0.36V  ^30  X  864  X0.367 


286  KINETIC  THEORY  OF  ENGINEERING  STRUCTURES        CHAP.  XIV 

DEFLECTION    POLYGONS    OF   HORIZONTAL    GIRDERS 
For  Conventional  Loading  of  One  Kip  Uniformly  Distributed  Over  Wetted  Length 


_*  £5.  <JLT.AN  T.  .  POAV  G  ON  . 

---  r  .W>TT«D*    LO.CTH       '-    17  FT.    "T  '  ~  '•  --- 


j 
DEFLECTION  ^OLYGON    FOR  GIRDER  I. 


M/I  POLYGON  FOR  GIRDERS  II  TO  V. 


DEFLECTION  "POLYftOIH  FOR  GIRDERS  II  TOY 


— h^a**C' 


All  deflections  are  20,000  times  actual 
measured  to  the  scale  of  lengths. 


FIGS.  68F. 


ART.  68 


MITERING  LOCK  GATES 


and  20,000  times  actual  in  feet  becomes 

^=0.00499(^5^)  =8.32  feet. 
\     iz     / 

This  crown  deflection  dc=8.32  ft.  is  now  incorporated  in  the  deflection  polygon 
for  Girder  I,J^ig.  68r,  by  laying  off  the  ordniate  C7C77=dc  and  then  drawing  the  "final 
closing  line  D'C"  to  complete  the  deflection  polygon.  The  ordinates  written  in  this 
diagram  represent  deflections  20,000  times  actual  in  feet. 

In  similar  manner  the  deflection  polygon  for  girders  II  to  V  is  drawn. 

9/  &•)  v  1 2 

The  actual  moment  deflection  at  point  6  is  now  "'  =0.00171  inch,  and  the 

20,000 

actual  deflection  due  to  shear  at  the  same  point  is 

&MR  5x2436 

d*  =2EF\  =  2 X29000 X 30 X0.375  =0-00019  inch> 

being  about  11  per  cent  of  that  due  to  moments.     Hence,  the  deflection  ordinates  are 
increased  11  per  cent  to  obtain  the  combined  moment  and  shear  deflection  polygon. 

The  crown  deflection  dc  is  found  as  before,  and  all  numerical  values  are  the  same 
except  the  cross-section  F  of  the  girder,  which  is  now  31.3  sq.in.  Hence 

1.4X292  1.4X24.8  .    ' 

c     29000  X  31.3  X  0.367 +30X864X0.367 

or  20,000  times  actual  in  feet  gives  dc=S.l5  feet. 

This  value  of  dc  is  now  used  to  complete  the  deflection  polygon  for  girders  II  to  V. 

The  average  deflection  dav  for  each  girder  is  now  computed  from  the  ordinates  taken 
from  the  deflection  polygons  Figs.  68F,  and  the  distances  Az  between  these  ordinates.  The 
values  Az  were  taken  in  inches  and  the  areas  finally  divided  by  the  length  Z  in  inches, 
which  eliminates  the  unit  of  length.  The  computations  are  given  in  the  following  table : 

TABLE  68c 
DEFLECTIONS  dav   FOR  THE   HORIZONTAL  GIRDERS. 


4z 

Horizontal  Girder  I. 

Horizontal  Girders  II  to  V. 

d* 

Areas 

d* 

Areas 

Inches. 

Feet. 

dJz. 

Feet. 

dJj. 

26.5 

0       +  1  .  58 

20.94 

0        +  1  .  50 

19.88 

27.5 

1.58+3.12 

64.63 

1.50+2.96 

61.33 

27.5 

3.12+4.64 

106.70 

2.96+4.38 

100.93 

31.2 

4.64  +  6.10 

167.54 

4.38+5.80 

158.81 

32.5 

6.10+7.30 

217.75 

5.80+6.92 

206.70 

32.4 

7.30+8.14 

250.13 

6.92+7.76 

237.82 

32.3 

8.14  +  8.52 

269.06 

7.76+8.20 

257.75 

31.8 

8.52  +  8.68 

273.48 

3.20+8.35 

263.15 

27.0 

8.68+8.62 

233.55 

8.35+8.30 

224.78 

26.3 

8.62  +  8.48 

224.86 

8.30+8.22 

217.24 

21.8 

8.48+8.32 

183.12 

8.22  +  8.15 

178.43 

316.8 

Tntn.1 

2011.76 

Total  

1926.82 

The  deflection  ordinates  d  are  20.000  times  actual  in  feet. 


288  KINETIC  THEORY  OF  ENGINEERING  STRUCTURES       CHAP.  XIV 

This  gives  for  horizontal  girder  I, 

2011.76          ..  „    6.35X12 

dav=  3168   =6.35  feet,  or  actually     2000Q--  0.00381  inch, 

and  for  girders  II  to  V, 


89  fi 

-  =6-082  feet,  or  actually  -  =0.00365  inch, 


which  values  will  be  used  in  Eqs.  (67o)  to  obtain  the  -deflections  d^   to  §5,  observing 
that  d'av  for  arches  II,  III,  IV  and  V  is  the  same  quantity. 

Attention  is  called  to  a  slight  oversight  on  Figs.  68F,  where  the  deflection  ordinates 
should  have  been  measured  at  the  numbered  points  2  to  10,  instead  of  on  the  lines  of 
the  w  loads.  The  effect  on  the  results  is  scarcely  appreciable  and  therefore,  the  error  was 
not  corrected. 

The  conventional  deflections  d  of  the  principal  system  are  determined  graphically  in  Figs. 
68c  and  68n,  for  the  case  of  full  contact  at  the  miter  sill.  This  requires  drawing  a  deflection 
polygon  for  the  principal  system  for  each  of  the  five  cases  of  conventional  loading. 

The  conventional  loading  for  the  horizontal  arches  was  made  equal  to  1  kip,  uni- 
formly distributed  over  the  wetted  length  of  a  gate  leaf,  being  1/1  =  1/27  =0.037  kip 
per  foot  of  arch.  Hence  the  same  load  \/l  =0.037  kip  must  be  employed  as  the  con- 
ventional load  for  the  principal  system. 

The  deflection  of  the  principal  system  is  made  up  of  the  deflection  of  the  vertical 
girder  and  the  elastic  displacements  of  the  two  determinate  supports  A  and  B.  For 
full  contact  at  the  miter  sill  the  B  support  is  assumed  as  being  immovable  and  the  A 
support  is  subject  to  the  elastic  displacement  of  the  top  horizontal  arch  caused  by  a  con- 
ventional load  X  =  1/1  acting  on  the  principal  system.  The  elastic  displacement  of  the 
point  A  is  determined  from  the  deflection  dav  just  found  for  the  top  horizontal  arch. 

For  the  condition  X\  —l/l,  there  are  no  moments  produced  in  any  part  of  the  vertical 
girder  and  hence  the  deflection  polygon  for  this  case  of  loading  would  show  no  bending 
effect  of  the  vertical  girder,  but  merely  an  elastic  displacement  of  the  A  support  which 
was  previously  found  to  be  d\-\=dav  for  the  top  horizontal  girder.  Hence,  this  con- 
ventional deflection  polygon  is  easily  drawn  and  becomes  a  triangle  aa'b,  Fig.  680,  wherein 
the  ordinate  aa'=dav-=  di_i  for  the  top  horizontal  girder.  This  deflection  line  also 
furnishes  the  values  dl_2,  ^1-3,  ^1-4  and  <?j_s,  which  are  respectively  equal  to  d2-i,  <V-i, 
^4_i  and  d$_i.  The  latter  are  the  end  ordinates  at  A  of  the  several  deflection  polygons 
to  be  drawn  for  the  conditions  X2  =  l/l  to  Xr,  =  \/l. 

Hence,  the  deflections  of  the  principal  system  are  easily  found  when  the  deflection 
lines  of  the  vertical  girder  are  once  drawn  for  each  of  the  several  conventional  loadings. 

The  deflection  lines  for  the  vertical  girder  are  drawn,  again  using  the  method  for 
plate  girders,  Art.  39.  However,  no  allowance  is  made  for  shear  because  the  web  is  too 
insignificant. 

A  set  of  w  loads  is  now  computed  for  each  of  the  loadings  Xz--=l/l  to  X5  =  l/l,  and 
observing  that  the  moment  of  inertia  7=971.6  in.4,  for  the  vertical  girder,  is  constant, 
it  is  preferable  to  make  these  loads  equal  to  MAx  for  a  pole  distance  H  =EI,  which  would 
give  actual  deflections  to  the  scale  of  lengths. 


AUT.  68 


MITERING  LOCK  GATES 


289 


For  simplicity  in  determining  the  moments  M  and  the  w  loads,  the  conventional 
loads  X  =  l  are  used,  giving  moments  27  times  too  large.  The  deflections  for  the  loads 
A7"  =  1/7  are  then  obtained  from 

MAx 

0  =^Tfr,  using  a  pole  distance 

Zt  Ji/l 

H  =27EI  =27  X29000  X971.6  =760,762,800, 

where  the  dimensions  are  kips  and  inches,  giving  actual  deflections  in  inches  to  the  scale 
of  lengths.  Instead,  however,  the  pole  was  made  #=27#//30,000  =25,360  for  deflec- 
tions 30,000  times  actual  to  the  scale  of  lengths,  and  w  loads  27  times  actual. 


w  loads  for  X2  =  1  Kip. 

w  loads  for  X3  —  1  Kip. 

Ar 

M  for  -X"2  =  l  Kip. 

w-MAx 

M  for  X3  =  l  Kip. 

w  =  MAx 

inches. 

Kips  and  inches. 

Kips  and  inches. 

' 

A2  =  0.2341       B.  =  0.7659 

43  =  0.4305      B3  =  0.5695 

Mt  =  0.7659  X     0.00=  0.0 

M!  =  O.O 

48.00 

w,=  882.0 

u>!=   655.2 

M2  =  0.2341X157.0  =36.75 

M2  =  0.5695  X   48.00=   27.3 

40.25 

w2  =  1289.6 

w2=  1561.7 

M3  =  0.2341  X  1  16.75  =  27.33 

M3  =  0.5695  X   88.25=   50.3 

40.25 

w>3=   910.5 

w3=  1674.4 

MI  =  0.2341  X   76.5   =17.91 

vl/4  =  0.4305=   76.50=   32.9 

40.25 

w4  =   539.4 

wt=   976.1 

M5  =  0.2341  X   36.25=   8.49 

M,  =  0.4305  X  36.25=   15.6 

36.25 

u>5=    153.9 

w>s=   282.8 

M6  =  0.2341X     0.0   =   0.0 

SM>  =3775.4 

M8  =  0.0 

~Lw  =  5150.2 

w  loads  for  X4  =  1  Kip. 

w  loads  for  Xb=  1  Kip. 

Ax 
inches. 

M  for  X4  =  l  Kip. 
Kips  and  inches. 

w  =  MAx 

MforX5=l  Kip. 
Kips  and  inches. 

„-** 

A4  =  0.627       £4  =  0.373 

/15  =  0.823       B&  =  0.177 

48  00 

M,  =  0.0 

Wl=  429.6 

1 

u>,=   204.0 

M2  =  0.373   X  48.0  =17.9 

w2  =  1022.4 

M2  =  0.177   X  48.00=  8.5 

w;2=   485.0 

M3  =  0.373   X  88.25X32.9 

w3  =  1626.1 

M3  =  0.177   X   88.25  =  15.6 

M>3=   770.8 

40.25 
36.25 

M4  =  0.373   X  128.5   =47.9 
M5  =  0.627   X  36.25  =  22.7 

w>4  =  1420.8 
M>5=  411.4 

M4  =  0.177   X  128.5  =22.7 
M5  =  0.177  X  168.75  =  29.9 

wt=  1058.6 

tws=  541.9 
Su>  =  3060.3 

290  KINETIC  THEORY  OF  ENGINEERING  STRUCTURES       CHAP.  XIV 

The  several  conventional  deflection  polygons,  for  loadings  X^l/l  =0.037  kip,  are  now 
drawn  as  illustrated  in  Figs.  68c  and  68n.  The  factor  30,000  was  introduced  to  obtain  dia- 
grams of  convenient  size  to  the  scale  of  the  drawing,  and  it  is  worthy  of  comment  that  the 
highest  desirable  accuracy  may  be  obtained  from  small  drawings  by  an  appropriate  choice 
of  the  scale  and  this  factor.  The  scale  employed  for  the  force  polygon  is  entirely  immate- 
rial so  long  as  it  is  convenient,  noting  that  the  w  loads  and  the  pole  must  be  measured 
to  the  same  scale.  The  pole  distance  remains  the  same  for  all  the  diagrams  and  is  25,360 
w  units. 

The  moment  diagrams  were  drawn  for  the  conventional  loadings  X  =  1  to  correspond 
with  the  moments  in  Table  68D,  and  serve  merely  to  locate  the  centers  of  gravity  gi,  <72, 
etc.,  which  are  the  points  of  application  for  the  w  loads. 

The  value  dav  =0.00381  inch,  as  previously  found  for  the  top  horizontal  arch,  now 
furnishes  <5i_1=30,000  dav  =  114.3  inches,  from  which  the  deflection  line  for  X^=l/l  is 
drawn.  This  diagram  then  furnishes  the  deflections  #1 -2=^2-1,  di_3-=ds_i,  ^1_4=o4_1 
and  ^1-5=^5-1,  which  are  necessary  in  fixing  the  closing  lines  of  the  several  other  deflec- 
tion polygons. 

With  this  explanation  the  drawings  should  be  readily  understandable  and  will 
furnish  all  the  deflections  d  required  in  solving  Eqs.  (67c) . 

The  problem  is  solved   completely  for  the  case  of  full  contact  at  the  miter  sill,   and 
the  method  of  deriving  the  deflections  for  the  case  of  no  sill  contact  is  shown  in  Figs.  68H. 

When  there  is  no  sill  contact  then  the  bottom  support  of  the  vertical  girder  becomes 
the  bottom  horizontal  arch  in  the  same  manner  as  the  principal  system  was  previously 
f ormed  for  the  top  support  at  A ,  requiring  now  the  determination  of  d'av  for  the  bottom 
arch  (for  a  loading  q  =0.037  k.  per  foot  of  arch)  as  was  done  for  the  other  arches.  This 
d'^  then  gives  <5e -6=30,000  d'av  for  the  principal  system  and  furnishes  the  means  of  draw- 
ing the  deflection  polygon  for  condition  X6  =  l/l  from  which  ^0-5=^5-6,  ^6-4=^4-6, 
e~tc.,  are  obtained.  These  deflections  then  serve  to  complete  the  conventional  deflection 
polygons  for  the  case  of  no  contact  at  the  miter  sill  in  the  manner  shown  for  the  two 
conditions  X4  =  l/l  and  X5  =  l/l  in  Figs.  68n.  Hence,  all  the  work  previously  accom- 
plished in  solving  the  case  for  full  contact  is  also  usable  for  the  case  of  no  contact,  though 
the  final  deflections  d  are  different  in  the  two  problems. 

The  formation  of  the  final  equations  for  the  solution  of  the  redundants  X  is  now  pos- 
sible by  introducing  the  numerical  values  for  all  the  deflections  d  into  Eqs.  (67c) ,  remem- 
bering that  these  were  all  taken  30,000  times  actual,  which  in  no  wise  interferes  with  their 
use  in  the  equations. 

The  numerical  terms  27>TO£l7n,  Zpmd2m,  etc.,  must  first  be  computed  for  the  actual 
water  loads  p  per  foot  of  the  horizontal  arches.  These  loads  are  given  on  the  diagram 
Fig.  68c  and  represent  the  actual  water  pressures  on  the  girders  for  vertical  depths  extending 
from  center  to  center  of  the  respective  spaces  between  arches.  Thus  the  load  pi  is  the 
water  pressure  per  horizontal  foot  of  arch  and  over  a  depth  extending  from  the  water 
surface  down  to  a  line  24  inches  below  the  top  arch.  The  load  p%  is  the  pressure  per 
horizontal  foot  of  the  arch  included  between  the  depths  38  inches  and  82.12  inches  below 
the  surface,  and  so  on  for  each  girder. 

The  double-subscript-bearing  deflections  are  taken  from  the  deflection  polygons  in 


ART.  08  MITERING  LOCK  GATES 

DEFLECTION  POLYGONS  FOR  THE  PRINCIPAL  SYSTEM 

FOR  FULL  CONTACT  AT  MITER  . 
THE  CONVENTIONAL  LOAD  *1/1=  1/27»O.O37  KIP  CON. 


291 


*  2    MOMENT   DjIAGRAM    FORjX^=IKIR 


All  deflections  are  30.000  times  actual  measured  to  the  scale  of  lengths. 
FlGS.  68G. 


292 


KINETIC  THEORY  OF  ENGINEERING  STRUCTURES        CHAP.  XIV 


DEFLECTION  POLYGONS  FOR  THE  PRINCIPAL  SYSTEM 

FOR   FULL  CONTACT  AT  MITER. 
THE  CONVENTIONAL  LOAD  =  1/1=  1/27=  O.037  KIP  CONCENTRATED. 


.   .    ._!.   .._, 

v 

upstream  chord. 

n 

Gravity  axis. 

t 

tn~ 

4>" 

1 

i 

2 

3 

4     r 

5 

L  AX=4ft^—  » 

I—     W-'iS  J 

»  4o'.'z5  > 

'  —     4o'/25  > 

<  —  36.25  —  • 

MOMENT   DIAGRAM    FOR|X  =  I 

I  21 


0» 

—  __j££Pi 
oi'o, 

nit 

°r 

i 

i 

MOMENT  DIAGRAM ^0^X5=  I  KIP 


I  DEFLECTION  POLYGON  FOR    Xft»1/l- 0.037 K-  NO  MITER  SILL  CONTACT. 


SCALES 


BO 
LOADS. 


1000   O  5000  10,000  \ 

All  deflections  ire  30,000  times  actual  measured  to  the  scale  of  lengths. 

FlGS.  68H. 


ART.  68 


MITERING  LOCK  GATES 


293 


Figs.  68a  to  H,  using  the  values  30,000  times  actual,  in  inches.  The  deflections  di  to  S5 
are  obtained  according  to  Eqs.  (67o),  using  30,000  times  the  deflection  dav,  found  for  the 
horizontal  arches. 

All  these  numerical  values  are  now  entered  in  Table  68E,  and  the  sum  products 
SpwAm  are  computed,  giving  all  required  data  for  writing  out  the  general  equations 
for  the  redundants. 

TABLE  68E 
COMPUTATION   OF  THE   TERMS    K  =  Zpmdam  FOR  EQS.  (67c) 


3  =  davX 

Pt. 

Pm 

sim 

Pm  "im 

°*m 

Pm3tm 

S3m 

Pm3vn 

34m 

Pm3<m 

ssm 

Pm*sm 

Eq.(67D) 

m 

Kips. 

1 

0.314 

114.3 

35.890 

87.6 

27.506 

65.2 

20.473 

42.8 

13.439 

20.2 

6.343 

114.3Xi 

9 

1.150 

87.6 

99.590 

69.6 

80.040 

53.8 

61.870 

37.0 

42.550 

18.4 

21.160 

109.  5Xi 

3 

1.783 

65.2 

116.252 

53.8 

95.925 

43.6 

77  .  739 

30.7 

54.738 

15.4 

27.458 

109.  5Xj 

4 

2.475 

42.8 

105.930 

37.0 

91.575 

30.7 

75.983 

22.0 

54.450 

11.0 

27.225 

109.5X4 

6 

3.015 

20.2 

60.903 

18.4 

55.476 

15.4 

46.431 

11.0 

33  .  165 

5.6 

16.884 

109.  5X. 

2Pm«im  = 

418.565 

*P»A«  = 

350.522 

*WW- 

282.496 

*Pm*m  = 

198.342 

**»*,«- 

99.070 

The  summations  cover  all  the  redundants  from  1  to  5  and  the  subscript  m  has  all  successive  values  from  1  to  5.     All 
the  values  d  are  30,000  times  actual  in  inches. 


FINAL  EQUATIONS  FOR  THE  REDUNDANTS 


Eq.  (1)     Xi^.j 
Eq.  (2)     X^,, 

Eq.  (3)     X^s-i 
etc. 


<?2_2  +  X3<52- 


(68B) 


etc. 


After  substituting  the  numerical  values  from  Table  68E,  the  following  equations 
are  obtained : 

(1)  114.3Xi  +87.6X2+&5.2X3+42.8X4+20.2X5  +  H4.aX1  =418.56 

(2)  87.6X!  +69.6X2  +53.8X3  +37.0X4  +  18.4X0  +  109.5X2  =350.52 

(3)  65.2X! +53.8X2 +43.6X3 +30.7X4  +  15.4X5 +  109.5X3=282.50  [     .     (68c) 

(4)  42.8Xi  +37.0X2  +30.7X3  +22.0X4  + 1  l.OXs  +  109.5X4  =  198.34 

(5)  20.2X! +18.4X2  +  15.4X3  +  H.OX4+  5.6X5  +  109.5X5=   99.07] 

The  terms  are  now  collected  in  the  above  equations  and  the  solution  is  carried  out 
in  the  following  tabular  forms: 


29t 


KINETIC  THEORY  OF  ENGINEERING  STRUCTURES,       CHAP.  XIV 
TABLE   68F— SOLUTION   OF  THE   EQUATIONS 


Operations. 

Coefficients  of  the  X'a. 

Numerical 
Term 
K. 

Xi 

Xi 

A'.i 

Xi 

X, 

Eq.  (1).. 

228.6 

87.6 

65.2 

42.8 

20.2 

418.56 

Eq.  (2)  

87.6 

179.1 
33.55 

53.8 
24.97 

37.0 
16.39 

18.4 
7.74 

350.52 
160.31 

Jg-frEq-  (l)]  =  0.383[Eq.  (1)] 

Eq.  (2),=Eq.  (2)-0.383[Eq.  (1)] 

145.55 

28.83 

20.61 

10.66 

190.21 

Eq.  (3)..                  65.2 

53.8 

153.1 

30.7 

15.4 

282.50 

65  2 
—  —  [Eq.  (l)]  =  0.285[Eq.  (1)] 

24.97 

18.58 

12.20 

5.76 

119.29 

Eq.  (3)!  =  Eq.  (3)-0.285[Eq.  (1)] 

28.83 

134.52 

18.50  ' 

9.64 

163.21 

9C  OQ 

^—  [Eq.  (2)1]  =  0.198[Eq.  (2),] 

5.71 

4.08 

2.11 

37.66 

Eq.  (3)2  =  Eq.  (3),  -0.198[Eq.  (2),] 

128.81 

14.42 

7.53 

125.55 

Eq.  (4)..                                              42.8 

37.0 

30.7 

131.5 

11.0 

198.34 

42  8 
—  —  [Eq.  (l)]  =  0.187[Eq.  (1)] 

16.39 

12.20 

8.00 

3.78 

78.27 

Eq.  (4),  =  Eq.  (4)-0.187[Eq(l)J 

20.61 

18.50 

123.50 

7.22 

120.07 

f)f\  f*~t 

^-5—  [Eq.  (2)J  =  0.142[Eq.  (2),] 

4.08 

2.92 

1.51 

27.01 

Eq.  (4)2  =  Eq.  (4)1-0.142[Eq.  (2)J 
14  42 

14.42 

120.58 

5.71 

93.06 

12881  "•    q'  (3)2-l-0-112LEcl-  (3)2-l 

.62 

.o4 

14  .Do 

Eq.  (4),  =  Eq.  (4)2-0.112[Eq.  (3)J 

118.96 

4.87 

79.00 

Eq.  (5)  20.2 

18.4 

15.4 

11.0 

115.1 

99.07 

20.2 
228^  [Eq.  (l)]  =  0.0884[Eq.  (1)] 

7.74 

5.76 

3.78 

-    1.78 

37.00 

Eq.  (5)1  =  Eq.  (5)-0.088[Eq.  (1)] 

10.66 

9.64 

7.22 

113.32 

62.07 

1  f\   f\f\ 

145.55  ^    q'  (^iJ-0-0732!^0!-  (2)J 

.  11 

.51 

.7o 

id.  yz 

Eq.  (5)2  =  Eq.  (5),-0.073[Eq.  (2)J 

7.53 

5.71 

112.54 

48.15 

12881^    q'  ^J-0-0585^-  (3):J 

.o4' 

.o4 

Eq.  (5)3  =  Eq.  (5)2-0.058[Eq.  (3)J 

4.87 

112.10 

40.81 

118%^    q'  ^3               "•    q'  ^  ^ 

Eq.  (5)4  =  Eq.  (5)3-0.04irEq.  (4)3] 

111.90 

37.58 

37.58 


ART.  68 


MITERING   LOCK    GATES 


295 


The  solution  of  the  equations  is  completely  given  in  Table  68r,  which  also  illustrates 
the  method.  The  successive  operations  are  expressed  in  the  first  column  in  unmis- 
takable language,  and  the  order  of  performing  the  computations  is  to  work  down  the 
table  first  entering  the  given  equations,  then  obtaining  the  products  /  [Eq.  (1)]  down 
through  each  sub-table,  and  finally  making  the  subtractions.  At  this  point  apply  the 
following  checks :  Note  that  the  symmetric  values  must  be  the  same  and  that  the  terms 
included  between  the  double-ruled  lines  and  those  in  the  columns  headed  K  are  not 
checked.  Therefore,  check  the  latter  terms  by  repeating  the  numerical  work  and  check 
the  symmetric  terms  by  inspection.  Then  proceed  with  the  work  of  obtaining  the 
products/"  [Eq.  (2)i]  and  repeat  the  same  programme  above  outlined. 

The  final  Eq.  (5)4  gives  the  value  of  ^5  =  0.3358  kip  per  horizontal  foot  of  horizontal 
girder  V.  The  other  redundants  are  then  determined  from  the  equations  in  the  last 
sub-table  of  Table  68F  by  successive  substitutions  as  indicated  in  the  following  Table  680. 


Terms. 

.sr, 

X, 

X3 

Xt 

X* 

K- 

99.07 

62.07 

48.15 

40.81 

37.58 

xtc> 

-38.65 

-38.02 

-37.79 

-37.64 

37.58 

xtct 

-  7.16 

-  4.70 

-  3.72 

-37.64 

x3cs 

-13.58 

-  8.50 

-41.51 

3.17 

=  0.3358 

x    3'17 

.AjCj 

—  18.73 

—  51.22 

.64 

4~4  87 

6  64 

-78.12 

10.85 

=  0.6509 

.A  •  — 

7.53 

10  85 

20.95 

=  0.8818 

2     10.66 

20.95 

=   1.0178 

1      20.2 

=  1.0371ki 

ps  per  horizontal 

"oot  of  gate. 

C  =  the  coefficients  from  the  last  of  Tables  68F. 

The  redundants  X  are  thus  expressed  in  kips  per  horizontal  foot  of  gate,  representing 
the  average  intensity  of  these  forces  for  the  average  vertical  stiffness  of  a  gate  leaf. 

The  reactions  A  and  B  and  the  moments  M  for  this  average  vertical  girder,  may  now 
be  found  from  Eqs.  (67e)  employing  the  values  in  the  following  table. 


296 


KINETIC  THEORY  OF  ENGINEERING  STRUCTURES        CHAP.  XIV 


TABLE  68n 
REACTIONS   AND   MOMENTS,   VERTICAL   GIRDER 


Point. 

p 
Kips  per  ft. 

X 
Kips  per  ft. 

p-X 
Kips  per  ft. 

h 
Inches. 

hi  -h 
Inches. 

(p-X)h 
Kip-inches. 

(p-X)(hi-h) 
Kip-inches. 

1 

0.314 

1.0371 

-0.7231 

205.0 

0.0 

-148.236 

0.0 

2 

1.150 

1.0178 

+  0.1322 

157.0 

48.0 

+   20.724 

6.336 

3 

1.783 

0.8818 

0.9012 

116.75 

88.25 

105.192 

79.513 

4 

2.475 

0.6509 

1.8241 

76.5 

128.5 

139.536 

234  .  380 

5 

3.015 

0.3358 

2.6792 

36.25 

168.75 

97.120 

452.115 

Totals 

8.737 

3  .  9234 

+  214.336 

772.344 

Eqs.  (67fi)  then  give  for  A,  B  and  M3 


779 
-h)  =i_' 


=3.767  kips. 


\ 


341.848  =  196.057  kip-inches, 


t      ,    My     196.057X23.4        __,_     . 
giving  for  f=—j—=  --  =-—  ~  -  =  4.727  kips  per  sq.m.  on  the  most  extreme  fiber  of 

J-  \j  i  J.  .O 

the  vertical  girder  at  the  panel  3. 

In  like  manner  the  stress  may  be  found  for  any  other  point  of  the  vertical  girder, 
using  the  results  given  in  Table  68n. 

B+PQ  represents  the  pressure  which  the  miter  sill   must  sustain  on  the  assumption 
of  full  contact. 

Finally  the  static  condition 

0      ././  -.     .     .     .     .     .     .      (680) 


must  be  fulfilled  by  the  above  numerical  results. 

The  deflections  of  the  vertical  girder  are  now  readily  found  from  Eqs.  (67c)  and  the 
redundants  X  which  are  now  known. 

The  deflections  dav  of  the  horizontal  girders  were  found  for  a  load  of  1  kip  uniformly 
distributed  over  the  wetted  length  of  27  ft.  Hence,  for  a  load  of  27.X"  uniformly  dis- 
tributed over  the  same  length,  the  deflection  would  be  27davX.  Accordingly  the 
deflections  d±  to  d$  become 

<?!  =27davXi  =27  X0.003S1  X  1.037  =0.107  in. 

d2  =27d'aVX2  =27  X0.00365  X  1.018  =0.100  in. 

33  =27d«vX3  =27  X0.00365  X0.882  =0.087  in. 

<?4  =27d'aVX4  =27  X  0.00365  X  0.651  =0.064  in. 

d5  =27d'm,X5  =27  X0.00365  X0.336  =0.033  in. 


ART.  68  MITERING  LOCK  GATES  297 

An  approximate  idea  of  the  deflection  curve  at  any  vertical  ra  other  than  the 
average,  may  be  obtained  by  assuming  that  the  same  proportionately  exists  between 
the  deflections  dav  and  dm  as  between  di  and  dm,  etc.,  where  the  point  ra  designates  any 
vertical  section  of  the  gate.  Thus  in  Fig.  68r,  for  a  vertical  girder  at  point  8,  the 
deflection  d8=8.68  ft.  for  the  top  horizontal  arch  while  dav  for  that  arch  was  6.35  ft., 
giving  a  ratio  d8/dav=  8.68/6.35  =  1.367.  For  the  arches  II  to  V  the  ratio  between 
similar  ordinates  becomes  dg/dap  =  8.35/6. 082  =  1.373. 

Hence,  in  considering  a  vertical  girder  at  point  8,  where  the  deflections  in  the 
horizontal  arches  are  about  37  per  cent  greater  than  the  average,  the  effect  on  all  the 
conventional  deflections  would  be  to  increase  them  approximately  by  the  same  constant 
percentage.  This  would  be  equivalent  to  multiplying  Eqs.  (68c)  by  a  constant  1.37 
and  since  the  water  loads  p  remain  unchanged  the  resulting  redundants  X  would  not 
be  materially  different  than  for  the  average  vertical  girder. 

However,  the  deflections  di  to  £5 would  be  increased  throughout  by  the  same  37  percent. 

Since  the  ratio  dm/dav  is  not  necessarily  constant,  though  nearly  so  in  the  present 
example,  this  proposition  cannot  be  accepted  for  more  than  it  is  worth,  and  approximate 
results  only  can  be  expected. 

Where  a  rigid  determination  is  desirable,  it  would  be  necessary  to  repeat  the  whole 
computation  for  the  actual  deflections  of  another  principal  system  with  vertical  girder 
at  point  8.  This  would  not  involve  very  much  labor,  however,  beyond  solving  Eqs. 
(68c)  for  a  new  set  of  numerical  coefficients. 

A  summary  of  the  steps  employed  in  the  solution  of  the  above  problem  -is  given  in 
closing  this  subject. 

Horizontal  arches.  Draw  a  moment  deflection  polygon  for  each  of  the  different 
arches,  using  a  conventional  load  of  one  kip  uniformly  distributed  over  the  wetted  length 
of  each  leaf.  These  deflections  are  then  combined  with  the  arch  deformation  for  axial 
thrust  or  rib  shortening  for  the  same  loading,  to  obtain  the  total  displacements  of  the 
several  points  of  the  arches  with  respect  to  the  fixed  quoin  post  supports. 

These  total  displacements  are  then  averaged  for  one  gate  leaf  by  dividing  the  area 
of  the  displacement  polygon  by  the  straight  length  of  the  girder,  giving  the  deflection 
dav  for  each  arch.  Then  d=davX,  and  for  the  top  arch  dav=dl.li  while  for  the  bottom 
arch  d'av=dnn  when  there  is  no  sill  contact  and  d'av=0  when  full  sill  contact  is  assumed. 

The  principal  system.  Draw  conventional  deflection  polygons  for  the  principal 
system,  using  conventional  loads  of  1  /I  kips  for  each  condition  X  =  l/l  where  I  is  the 
wetted  length  of  one  gate  leaf  as  above.  This  gives  the  conventional  deflections  of  all 
panel  points  of  the  vertical  girder  for  each  condition  of  loading,  remembering  that  the 
vertical  girder  represents  a  strip  of  gate  one  foot  thick  and  of  average  cross-section. 

The  water  loads  p  per  horizontal  foot  of  gate  for  each  arch,  are  then  used  as  the 
actual  loads  in  computing  the  numerical  terms  Spm8am  for  each  condition  X-l/l.   _ 
resulting  values  X  are  then  the  redundant  reactions  of  the  average  vertical  gir 
against  the  horizontal  arches  per  linear  foot  of  the  latter. 

When  the  vertical  girder  is  a  truss  frame  instead  of  a  plate  girder,  then  the  method 
of  deflections  given  in  Art.  36  must  be  used,  neglecting  the  effect  of  the  web  member* 
as  the  nature  of  the  problem  does  not  warrant  the  extreme  accuracy  impl* 
web  system  is  considered. 


CHAPTER  XV 
FIXED   MASONRY  ARCHES 

ART.  69.     GENERAL   CONSIDERATIONS 

The  fixed  arch  is  statically  indeterminate  in  the  third  degree,  or,  as  commonly 
expressed,  it  involves  three  redundant  conditions,  provided  the  arch  ring  may  be  treated 
as  an  elastic  body. 

According  to  the  theory  of  elastic  deformation,  these  redundant  conditions  are 
determined  from  the  elastic  properties  of  the  material,  while  for  the  older  "  line  of 
thrust "  method  the  redundants  are  merely  approximated  by  assigning  certain  assumed 
conditions  to  be  fulfilled  by  the  line  of  thrust. 

A  line  of  thrust  is  defined  as  an  equilibrium  polygon  for  any  case  of  simultaneous 
loads  acting  on  the  arch  ring.  Hence,  for  any  section,  the  adjacent  elements  of  the 
line  of  thrust  will  represent  the  resultant  of  all  the  external  forces  on  either  side  of  the 
section,  including  the  reactions..  From  this  it  follows  that  the  sum  of  the  moments  of 
all  external  forces  about  any  point  on  the  line  of  thrust,  or  resultant  polygon,  must  always 
be  zero. 

Therefore,  if  the  external  redundant  reactions  were  known,  then  the  real  resultant 
polygon  could  be  drawn,  but  since  the  redundants  can  be  found  only  from  the  theory 
of  elasticity,  the  line  of  thrust  method  fails  to  give  a  direct  solution. 

A  resultant  polygon  may  always  be  drawn  through  three  given  points,  as  will 
be  shown  later.  Hence,  by  successive  trials,  a  resultant  polygon  may  be  approximated 
so  as  to  fulfill  certain  requirements  which  are  now  discussed.  The  line  of  thrust  is 
henceforth  called  the  resultant  polygon,  this  being  a  more  appropriate  designation. 

The  theory  of  elasticity  fixes  the  position  of  the  resultant  polygon  in  terms  of  the  modu- 
lus E  of  the  material  and  the  elastic  deformation  of  the  arch  ring.  According  to  the 
older  method  an  infinite  number  of  resultant  polygons  may  be  constructed  for  the  same 
arch  and  same  loading,  and  it  becomes  a  question  to  decide  which  of  all  possible  lines 
is  the  most  probable. 

Hagen  (1844  and  1862),  according  to  his  "  theory  of  the  most  favorable  distribution 
of  stress,"  defines  the  most  probable  resultant  polygon  as  the  one  for  which  the  vertical 
projections  of  the  minimum  distances,  between  the  sides  of  this  polygon  and  the  bound- 
ing lines  of  the  arch  ring,  become  equal. 

Culmann  (1866)  advanced  the  theory  that  of  all  the  possible  resultant  polygons, 
the  most  probable  one  must  approach  the  arch  center  line  in  such  manner  as  to  reduce 
the  stresses  in  the  critical  sections  to  a  minimum.  Carvallo  (1853)  and  Durand-Claye 
(1867)  adopted  this  theory  with  slight  modifications. 


ART.  69  FIXED  MASONRY  ARCHES  299 

The  theory  of  elastic  deformation  was  introduced  by  Navier  in  1826,  by  his  analysis 
of  the  stresses  on  an  arch  section,  in  which  he  assumes  a  combined  thrust  and  bending 
moment  distributed  over  the  section.  According  to  Navier,  the  stress  on  the  extreme 
fiber  of  any  arch  section  becomes 

f_N     My_N    M 

J  ~^  ......... 


where  N  is  the  normal  thrust  on  the  section,  M  the  bending  moment,  F  the  area,  y  the 
distance  from  the  neutral  axis  to  the  extreme  fiber  and  /  and  W  the  moments  of  inertia 
and  of  resistance,  respectively,  of  the  cross-section.  The  formula  when  applied  to  curved 
beams,  or  arches,  is  not  strictly  accurate,  but  when  the  radius  of  curvature  is  large  in  com- 
parison with  the  depth  of  the  arch  ring,  a  condition  which  always  obtains  in  ordinary 
arches,  then  the  error  due  to  omitting  curvature  is  extremely  small  and  is  never  considered. 

The  experiments  of  Bauschinger  (1834-93),  Koepcke  (1877),  and  the  Austrian  Society 
of  Engineers  and  Architects  (1895),  have  substantiated  the  correctness  of  the  theory  of 
elasticity  within  knowable  limits  of  the  elastic  properties  of  building  materials. 

Nearly  all  modern  writers  on  this  subject  adopt  Navier's  law  as  the  basis  of  investiga- 
tion. The  various  arch  theories  by  Winkler  (1867),  Mohr  (1870),  Belpaire  (1877),  J. 
Weyrauch  (1878),  F.  Engesser  (1880),  Mueller-Breslau  (1880),  and  those  of  American 
writers,  differ  principally  in  manner  of  presenting  the  subject.  Each  author  has  con- 
tributed something  in  the  direction  of  rendering  the  theory  of  elasticity  more  usable,  and 
this  is  especially  desirable  even  at  the  present  time. 

Professor  Winkler  was  the  first  to  prove  that  the  most  favorable  resultant  polygon  in  an 
unsymmetric  arch  must  intersect  the  arch  center  line  in  at  least  three  points,  while  -for 
symmetric  arches  four  such  intersections  are  necessary.  He  also,  proved  that  according 
to  the  theory  of  elasticity  the  most  probable  resultant  polygon  is  the  one  for  which  the 
I'esidual  departures  from  the  arch  center  line  become  a  minimum,  according  to  the  method 
of  least  squares.  This  then  furnishes  the  real  criterion  with  which  to  judge  an  arch  design 
in  connection  with  certain  other  limitations  to  be  discussed  later. 

The  common  graphic  solution  is  nothing  more  than  the  application  of  the  principles 
of  the  three-hinged  arch  to  the  fixed  arch  by  assuming  the  location  of  the  hinged  points, 
at  the  crown  and  springing.  The  location  of  these  hinged  points  may  be  altered  until 
such  a  resultant  polygon  is  found  as  will  approach  the  arch  center  line  in  the  manner 
required  by  the  theory  of  elasticity. 

The  most  probable  resultant  polygon,  when  found,  must  remain  within  the  middle 
third  of  the  ring,  when  tensile  stress  is  not  permissible.  This  follows  from  Eq.  (69A). 
Hence  it  is  clear  that  the  most  economic  masonry  arch  is  one  for  which  the  center  line 
coincides  most  closely  with  the  resultant  polygon  drawn  for  the  case  of  average  or  half 
total  live  load,  provided  the  arch  ring  has  sufficient  thickness  at  all  points  to  prevent 
excessive  unit  stresses  due  to  unsymmetric  loading.  This  idea  will  be  followec 
treating  of  methods  for  preliminary  designs. 

Formerlv  an  arch  problem  was  considered  solved,  when,  for  a  given  span,  rise  and 
loading  a  resultant  polygon  could  be  found  which,  for  the  critical  position  of  the  live 
load,  remained  entirely  within  the  middle  third  of  the  ring.  However,  this  is  far  from 


300  KINETIC  THEORY  OF  ENGINEERING  STRUCTURES          CHAP.  XV 

constituting  an  acceptable  solution,  as  will  be  shown  in  the  following.  The  fact  that  most 
arches  have  stood  so  well  is  no  defense  of  this  method  of  designing,  especially  when  it 
is  understood  that  the  usual  factor  of  safety  employed  in  such  designs  ranges  between 
ten  and  sixty,  rarely  going  below  twenty. 

Furthermore,  solid  spandrel  arches  are  usually  designed  with  very  high  safety 
without  considering  the  additional  carrying  capacity  of  the  masonry  in  the  spandrel, 
which  is  often  sufficient  to  sustain  the  loads  without  regard  to  the  ring. 

The  author  adds  the  criticism  that  few  fixed  arches  exist  which  are  not  disfigured  by 
unsightly  cracks  and  this,  in  conjunction  with  the  very  wasteful  expenditure  of  mate- 
rial necessitated  by  low  unit  stresses,  constitutes  a  fertile  field  for  doubt  regarding  the 
expediency  and  justification  of  building  fixed  arches  in  this  progressive  age. 

While  it  is  freely  admitted  that  these  masonry  structures,  many  of  which  are  monu- 
mental in  character,  have  stood  well,  and  will  undoubtedly  continue  to  stand,  every 
engineer  must  agree  that  a  cracked  arch  is  neither  an  achievement  nor 'the  ultimate  aim 
of  his  ambition,  to  say  nothing  of  the  dissatisfaction  which  such  a  blemish  constitutes 
to  both  engineer  and  owner. 

Practically,  the  objections  to  fixed  masonry  arches  may  seem  trivial  and  mostly 
a  matter  of  sentiment,  while  theoretically  they  are  important.  The  remedy  for  all  these 
objections  is  to  introduce  hinged  joints  during  construction  or  preferably  for  all  time. 
This  has  been  practiced  in  Germany  since  1880,  and  has  given  excellent  satisfaction, 
though  many  fixed  arches  are  still  being  built.  This  will  continue  to  be  the  case  so  long 
as  we  do  not  come  to  recognize  that  "  new  truths  are  better  than  old  errors."  * 

The  theory  ot  elasticity  is  most  applicable  to  arches  of  small  depth  of  ring  and  high 
rise,  and  the  analysis  of  stresses  will  be  more  accurate  the  nearer  the  shape  of  the  arch 
center  line  coincides  with  the  resultant  polygon  for  average  loading. 

This  is  true  when  dealing  with  the  arch  ring  alone,  as  the  ring  is  the  only  portion  of 
an  arch  which  could  be  expected  to  follow  the  laws  of  elastic  deformation.  Hence,  in  order 
that  the  theory  shall  be  at  all  applicable,  the  relation  between  the  arch  ring  and  its  span- 
drel filling  must  be  such  as  to  allow  the  ring  perfect  freedom  to  undergo  elastic 
deformations. 

The  style  of  structure  which  best  satisfies  this  requirement  is  one  where  the  arch 
ring  supports  the  roadway  on  a  succession  of  piers,  producing  what  is  called  an  open 
spandrel  filling.  Even  here  the  roadway  should  be  provided  with  expansion  at  one  end. 

Any  other  style  of  spandrel  will  introduce  further  redundants  which  are  wholly 
beyond  analysis,  because  the  spandrel  is  then  subjected  to  bending  and  direct  stress 
the  samfe  as  the  ring1. 

Therefore,  to  apply  the  theory  of  elasticity  or  in  fact  any  theory,  to  a  solid  spandrel 
fixed  arch  is,  in  the  author's  opinion,  a  mere  waste  of  time  which  should  be  expended  in 
a  more  profitable  pursuit.  In  the  following,  nothing  but  open  spandrel  arch  bridges 
are  dealt  with. 

While  the  theory  of  elasticity  is  Undoubtedly  the  only  trustworthy  basis  for  the 

*  See  paper  by  the  author  on  Three-Hinged  Masonry  Arches,  Trans.  Am.  Soc.  C.  E.,  Vol.  XL,  1898, 
p.  31. 


ART-  70  FIXED  MASONRY  ARCHES  301 

analysis  of  fixed  arches,  and  should  always  be  employed  in  designing  structures  of  any 
importance,  it  does  not  follow  that  the  older  methods  of  investigation,  and  many  of  the 
empiric  rules  in  common  use,  are  all  worthless.  On  the  contrary,  the  preliminary  design 
can  be  carried  out  most  efficiently  by  the  aid  of  these  simpler  methods,  and  the  final 
design  should  then  be  tested  by  the  application  of  the  more  exact  method  of  the 
theory  of  elasticity. 

The  theory  of  fixed  arches  has  reached  a  status  of  perfection  quite  in  keeping  with 
the  nature  of  the  problem,  the  still  existing  uncertainty  being  a  function  of  the  material 
and  other  circumstances,  depending  on  the  rigidity  of  the  abutment  foundations,  condi- 
tions of  erection,  etc.,  all  of  which  can  never  be  definitely  known  nor  be  entirely  eliminated 
in  the  best  designs. 

When  the  loading  is  not  known  with  considerable  accuracy,  as  for  arches,  sustain- 
ing high  earth  banks,  the  design  must  necessarily  remain  more  or  less  indefinite,  though 
the  elastic  property  of  the  superimposed  earth  assists  greatly  in  distributing  pressures 
and  rendering  conditions  more  favorable.  Also  in  dealing  with  small  arches,  such  as  cul- 
verts and  road  crossings,  the  resultant  polygon  theory  will  probably  continue  to  be  the 
only  method  of  approximate  analysis. 

ART.  70.     MODERN  METHODS   OF   CONSTRUCTION 

Many  difficulties  are  encountered  in  the  construction  of  fixed  masonry  arches,  owing 
particularly  to  insufficient  elasticity  in  masonry.  The  natural  deformations  in  the  arch, 
caused  by  shrinkage  of  the  masonry  or  concrete,  due  to  the  setting  process,  stress  and 
temperature,  usually  cause  cracks,  which,  while  rarely  of  a  serious  character,  are  reasons 
for  discouragement  to  the  engineer,  who  has  probably  applied  eveTy  known  precaution 
to  prevent  their  occurrence. 

So  long  as  there  are  no  abutment  displacements  after  completion  of  an  arch,  the 
above  difficulties  may  be  fairly  well  controlled.  However,  when  an  arch  was  designed 
for  certain  allowable  unit  stress  on  the  extreme  fiber  at  the  critical  points,  then,  if  the 
structure  is  to  be  regarded  safe  for  all  time,  the  original  stress  must  not  be  exceeded  even 
when  the  material  increases  in  strength  (as  by  setting  of  the  cemjent)  or  when  the  arch 
or  its  abutments  undergo  slight  elastic  or  permanent  displacements.  For  this  there 
is  absolutely  no  assurance,  though  it  is  an  essential  necessity  in  the  general  assumption. 

If  an  arch  ring  could  be  built  in  such  manner  that  its  resultant  polygon  would  pass 
through  the  center  of  the  ring  at  the  crown  and  springing  points  at  the  time  of  releasing 
the  falsework,  a  large  proportion  of  the  redundant  stress  could  be  prevented.  In  other 
words,  the  bending  moments  at  the  critical  points  would  then  be  almost  zero  for  sym- 
metric loading,  reserving  the  strength  for  the  unsymmetric  live  load  and  other  contingencies 
affecting  the  shape  of  the  arch  ring. 

By  the  ordinary  process  of  construction,  the  arch  is  commenced  at  the  abutments 
and  the  load  is  gradually  applied  to  the  falsework,  thus  distorting  the  latter  and  causing 
bending  stresses  in  the  completed  portion  of  the  ring.  The  final  settling  of  the  ring, 
produced  by  the  contraction  of  the  mortar  or  concrete,  as  a  result  of  compression  and 
shrinkage  during  setting,  even  when  the  abutments  remain  immovable,  are  sufficient 


302  KINETIC  THEORY  OF  ENGINEERING  STRUCTURES  CHAP.  XV 

to  create  serious  initial  stresses,  thus  destroying  to  a  large  extent  the  usefulness  of  the 
structure  and  frequently  cracking  the  arch. 

Several  methods  have  come  into  use  by  which  these  initial  arch  stresses  may  be  more 
or  less  completely  prevented.  One  of  the  oldest  of  these  consisted  in  setting  the  arch 
stones  on  the  entire  falsework,  spacing  the  joints  by  small  strips  of  wood,  and  lastly 
filling  all  the  joints  simultaneously  with  mortar.  This  is  still  a  very  good  programme  for 
construction  of  small  arches,  but  not  so  well  adapted  to  large  structures,  owing  to  the 
excessive  stresses  produced  in  the  falsework,  frequently  causing  considerable  settlement 
before  the  ring  can  be  closed.  However,  the  falsework  could  be  wedged  up  just  previous 
to  filling  the  joints. 

Another  method,  frequently  employed  in  the  construction  of  brick  and  concrete 
arches,  is  to  close  a  complete  ring  adjacent  to  the  falsework,  and  after  allowing  this 
course  to  set  firmly,  the  remainder  of  the  arch  is  completed,  thus  carrying  the  load  by 
the  first  ring  rather  than  by  the  falsework.  By  this  means  the  settlement  during 
construction  becomes  very  slight,  but,  as  is  readily  seen,  the  intrados  is  excessively 
stressed,  a  condition  which  cannot  be  averted  and  which  is  highly  objectionable. 

A  programme  of  construction  frequently  followed  on  large  arches  is  to  commence  work 
simultaneously  at  two,  four  or  six  points  of  the  ring  and  closing  at  three,  five  or  seven 
points,  respectively.  The  larger  the  span  the  greater  the  number  of  points  of  commence- 
ment. 

The  most  modern  method  consists  of  the  introduction  of  temporary  flexible  or 
hinged  joints  at  the  crown  and  springing  and  following  the  last-mentioned  programme 
of  construction.  These  flexible  joints  are  made  of  stone  with  curved  roller-like  surfaces; 
or  iron  blocks  may  be  used.  By  this  means  the  resultant  polygon  becomes  determinate 
during  the  period  of  construction.  After  the  falsework  is  removed  and  the  structure 
has  assumed  a  normal  condition  of  stress,  these  open  joints  are  grouted  with  cement 
mortar. 

Sheet  lead  or  lead  blocks  have  also  been  used  for  this  purpose,  allowing  sufficient 
surface  of  contact  on  the  arch  center  line  to  carry  the  pressure  without  causing  the  lead 
to  flow.  A  safe  allowable  pressure  for  lead  is  from  1000  to  1500  Ibs.  per  sq.  in. 

The  author  would  recommend  an  alloy  of  lead  with  from  5  to  10  per  cent  of  copper 
added,  thus  permitting  an  allowable  unit  stress  of  about  2500  Ibs.  per  sq.  in.  The 
last  mentioned  method  gives  very  good  results  in  preventing  cracks  and  excessive 
stresses,  during  construction,  but  subsequent  abutment  displacements  or  other  changes 
cannot  be  compensated  and  still  remain  as  a  serious  objection  to  this  class  of  structure. 
Eventual  settlements  in  foundation  masonry,  contraction  of  mortar  or  concrete  due 
to  setting  and  drying  out;  compression  due  to  loading,  elastic  deformations  caused  by 
temperature  and  load  effects — all  these  influences  are  ever  present  to  create  stresses 
which,  in  spite  of  all  precautions  during  construction,  may  attain  dangerous  proportions 
and  make  it  utterly  impossible  to  estimate  the  ultimate  strength  of  a  fixed  masonry 
arch. 

However,  it  should  be  remembered  that,  for  masonry  arches,  the  live  load  is  generally 
a  small  fraction  of  the  dead  load,  and  for  this  reason  an  arch  which  is  sufficiently  strong 
to  sustain  its  own  weight  permanently  will  carry  temporary  live  loads  with  perfect  safety. 


ART.  70  FIXED  MASONRY  ARCHES 

Masonry  arches  possess  the  redeeming  feature  that  when  stressed  to  the  breaking 
limit  the  masonry  generally  chips  near  the  surface,  thus  relieving  the  stress  and  allowing 
the  resultant  polygon  to  return  to  a  more  favorable  position.  In  this  manner  an  arch 
which  has  become  distorted  may  readjust  itself  to  a  new  condition  of  stress. 

Hence  fixed  masonry  arches,  in  view  of  their  past  history,  cannot  be  condemned 
on  account  of  insufficient  safety  which  they  afford,  but  because  masonry  is  not  a  suitable 
material  from  the  criterion  of  inability  to  withstand  elastic  deformations  well.  A  fixed 
plate  girder  arch  would  be  a  far  more  rational  structure  than  a  fixed  masonry  arch,  though 
the  former  would  require  frequent  painting  and  even  then  might  deteriorate  and  thus 
outlive  its  usefulness  earlier  than  the  masonry  structure.  But  from  an  engineering 
standpoint  the  plate  girder  would  probably  be  more  satisfactory  because  it  would  not 
open  up  cracks  and  thus  bring  discredit  to  the  designer  and  builder. 

In  this  connection  it  may  be  of  interest  to  recall  the  recommendations  proposed  by 
the  Austrian  Society  of  Engineers  and  Architects,  which,  if  strictly  carried  out,  would 
limit  the  application  of  fixed  masonry  arches  to  comparatively  short  spans  with  bed 
rock  foundations.  These  recommendations  require  the  fulfillment  of  the  following 
conditions : 

1.  The  abutments  must  be  perfectly  rigid. 

2.  The  falsework  must   retain  its   form   during   the  period  of  construction  of  the 
arch  ring. 

3.  The  masonry   must  be  of  the  best  quality. 

4.  The  construction  of  the  arch  ring  must  be  most  carefully  conducted. 

5.  The  falsework  must  not  be  released  until  the  cement  has  thoroughly  set. 

6.  When  the  falsework  is  finally  released,  it  must  be  done  gradually  and  uniformly. 

7.  To  this  the  author  adds  that  the  arch  ring  should  be  closed  at  the  lowest  possible 
temperature. 

The  use  of  sand  jacks  or  sand  pots,  which  was  introduced  in  1854  during  the  construc- 
tion of  the  Austerlitz  Bridge,  in  Paris,  offers  a  very  novel  and  efficient  means  of  releasing 
falsework. 

The  necessity  of  these  recommendations  is  clearly  understood  after  what  has  been 
said  regarding  the  theory  of  fixed  masonry  arches  and  its  practical  applications  and 
limitations. 

The  two  first  conditions  can  be  realized  only  when  rock  foundations  are  available. 
The  other  requirements  can  generally  be  fulfilled  by  exercising  proper  care  and  by 
permitting  nothing  other  than  first-class  materials  and  workmanship. 

From  the  above  review  of  the  subject  and  actuated  by  many  years  of  practical 
experience,  the  author  ventures  the  following  opinion  as  to  the  general  advisability  of 
adopting  the  fixed  type  of  masonry  arch  for  any  particular  case  in  hand:  If  the  span 
i  is  moderate  and  the  rise  h  comparatively  high,  such  that  l/h  will  be  between  two  and 
four,  then  for  bed  rock  foundation  and  open  spandrels  the  fixed  masonry  arch 'may  be 
chosen  as  an  appropriate  type  of  structure.  The  design  for  such  a  bridge  should  be 
based  on  a  factor  of  safety  of  at  least  ten,  allowing  no  tension  unless  carried  by  steel 
reinforcement,  and  after  the  preliminary  design  is  completed  it  should  receive  a  final 
analysis  according  to  the  theory  of  elasticity. 


304  KINETIC  THEORY  OF  ENGINEERING  STRUCTURES          CHAP.  XV 

When  all  this  is  followed  by  a  most  careful  and  approved  system  of  construction, 
it  may  frequently  happen  that  the  results  will  not  prove  entirely  satisfactory  to  the 
engineer.  However,  such  a  structure  will  be  certain  to  last  well,  to  cost  little  or  nothing 
for  maintenance,  and  to  be  safe  practically  for  all  time,  even  though  small  cracks  may 
appear  at  any  time  after  removing  the  falsework. 

On  the  other  hand,  when  the  spandrel  filling  must  for  some  reason  be  made  solid. 
and  bed  rock  foundations  are  not  available,  also  when  the  span  is  great  and  the  rise 
small,  a  fixed  masonry  arch  should  never  be  built.  A  hinged  arch  is  the  only  rational 
solution  for  such  a  case  in  the  light  of  modern  engineering  experience,  and  that  type 
can  be  employed  only  when  the  spandrels  can  be  so  arranged  as  to  allow  of  proper 
expansion  or  rotation  at  the  hinged  points. 

All  this  applies  strictly  to  bridges  and  riot  to  small  arched  culverts  and  floors  where 
the  hinged  type  is  in  many  ways  impracticable  and  the  fixed  type,  preferably  of  reinforced 
concrete,  has  given  good  satisfaction. 

It  is  thus  seen  that  the  successful  construction  of  a  large  masonry  arch  may  justly 
be  considered  a  masterpiece  of  engineering  achievement  and  requires  far  more  intimate 
knowledge  and  experience  than  is  required  for  a  steel  structure  of  the  same  span. 

Masonry  and  concrete  are  materials  which  are  solely  adapted  to  carrying  compressive 
stress.  Hence,  it  is  rational  to  reinforce  this  material  to  better  resist  tensile  stress  by 
the  introduction  of  steel,  but  this  should  be  confined  within  proper  limits  and  with  the 
sole  aim  of  developing  thereby  the  otherwise  latent  capacity  of  the  material  to  do  work 
in  compression. 

When,  therefore,  steel  is  used  to  reinforce  concrete  in  compression,  the  result  is 
bound  to  be  wasteful,  unprecedented,  and  a  monument  to  engineering  misconception. 
Designs  of  this  kind  should  be  discountenanced,  both  from  an  aesthetic  as  well  as  from 
an  engineering  standpoint,  because  they  imply  a  misapplication  of  the  true  purpose  of 
both  masonry  and  steel. 

ART.  71.     PRELIMINARY   DESIGNS 

On  the  supposition  that  a  fixed  masonry  arch  is  feasible  only  when  the  foundations 
are  rigid,  it  is  fair  to  assume  that  the  fixed  conditions  will  remain  unchanged  even 
though  the  transmission  of  stress  may  call  forth  very  slight  elastic  changes  in  the  abut- 
ments. 

In  any  problem  the  given  data  are  usually  the  live  load  to  be  carried,  the  clear  span 
and  the  clear  rise  of  the  intrados  and  the  allowable  unit  stress.  The  elevation  of  the 
roadway  is  also  known. 

The  preliminary  design  would  then  include  a  reasonably  accurate  determination  of 
the  thickness  and  shape  of  the  arch  ring  at  all  points  over  the  span. 

To  accomplish  this,  it  is  necessary  first  to  decide  on  the  general  outlines  of  the 
bridge,  particularly  on  the  design  of  the  spandrel  filling  and  the  roadway,  which  should  be 
completely  detailed  before  proceeding  to  the  design  of  the  arch  ring.  By  this  means 
a  large  portion  of  the  dead  load  can  be  accurately  estimated,  leaving  the  bare  arch  ring 
as  the  only  variable  portion  of  the  dead  load. 


ART-71  FIXED  MASONRY  ARCHES 


305 


Since  the  dead  load  for  masonry  arches  is  usually  large  compared  with  the  live 
load  this  procedure  is  clearly  indicated. 

The  next  step  should  be  to  determine  the  axis  of  the  arch  ring  so  that  it  will 
coincide  with  the  resultant  polygon  for  an  average  load,  which  latter  is  taken  as  the 
dead  load  plus  half  the  uniformly  distributed  live  load.  In  preliminary  designs  the  live 
load,  whatever  it  may  be,  is  always  taken  as  uniformly  distributed. 

The  shape  of  the  arch  center  line  thus  obtained  will  be  the  one  yielding  the  most 
economic  sections  for  the  unsymmetric  maximum  and  minimum  live  loads. 

However,  this  center  line  cannot  be  directly  found  because  the  resultant  polygon 
upon  which  it  depends  must  be  determined  before  the  dimensions  of  the  arch  ring  and 
its  weight  are  definitely  known. 

Hence,  this  part  of  the  problem  must  be  solved  by  successive  approximations  and 
trials.  We  must  seek  to  find  such  an  arch  ring  for  which  the  resultant  polygon  will 
coincide  approximately  with  the  center  line. 

With  the  aid  of  approximate  formulae,  the  thicknesses  of  the  arch  ring  at  the  crown 
and  abutments  are  determined,  and  from  these  the  radius  of  curvature  at  the  crown  is 
approximated  to  find  the  first  arch  center  line.  The  weight  of  this  arch  ring  is  now 
estimated  and  combined  with  the  weight  of  roadway  arid  half  live  load,  and  a  resultant 
polygon  is  then  drawn  through  the  centers  of  the  crown  and  springing  sections.  This 
then  gives  a  criterion  as  to  the  modifications  required  to  make  a  second  approximation 
for  a  correct  center  line.  This  process  is  continued  until  a  center  line  is  found  which 
coincides  closely  with  the  resultant  polygon. 

The  maximum  stress  for  unsymmetric  loading  is  now  estimated  from  another 
resultant  polygon  drawn  for  the  critical  quarter  points  and  this  then  furnishes  the 
criterion  for  the  thickness  of  the  arch  ring.  If  this  stress  is  sufficiently  close  to  the 
allowable  stress,  the  preliminary  design  may  be  accepted  and  the  final  rigid  analysis 
may  then  be  undertaken.  Otherwise  further  alterations  in  the  dimensions  must  be 
made  until  the  preliminary  design  is  sufficiently  close  to  the  assigned  requirements. 

The  approximate  formulae  and  method  of  conducting  this  investigation  will  now 
be  given,  and  this  will  constitute  a  complete  solution  as  far  as  is  possible  by  the  use  of 
the  older,  resultant  polygon  method. 

Fig.  7lA  represents  a  symmetric  arch  ring  of  unit  thickness.  If  the  arch  were 
slightly  unsymmetric,  the  present  solution  might  answer  for  the  symmetric  portion  and 
the  resultant  polygon  could  afterward  be  extended  into  the  unsymmetric  portion  which 
was  omitted  in  the  first  analysis.  However,  this  could  not  be  done  in  the  final  rigid 
analysis  by  the  theory  of  elasticity  and  should  never  be  considered  for  badly  unsymmetric 
structures. 

All  pressures  and  loads  will  be  expressed  in  terms  of  cubic  feet  of  masonry  per  square 
foot  of  surface,  so  that  all  areas  on  the  drawing  icpresent  true  relations  of  loads  from 
which  actual  weights  are  obtained  by  multiplying  the  areas  by  the  weight  of  one  cubic 
foot  of  masonry.  This  is  the  most  convenient  unit  for  graphic  solutions,  as  it  does  away 
with  estimating  the  weight  of  the  arch  ring  and  spandrel  filling. 

The  following  data  are  supposed  to  be  given :  The  system  of  loading;  width  of  roadway ; 
span  and  rise  of  intrados;  and  the  allowable  unit  stresses  of  the  masonry  or  concrete.  , 


306 


KINETIC  THEORY  OF  ENGINEERING  STRUCTURES          CHAP.  XV 


The  first  step  should  be  to  sketch  a  general  outline  of  the  structure  in  sufficient 
detail  to  design  the  roadway  and  the  spandrel  filling.  This  will  be  considered  completed 
and  will  not  be  discussed  here. 

If  the  bridge  is  to  be  for  a  single  track  railway  for  Cooper's  E  60  loading,  span  150 
ft.,  rise  50  ft.,  and  width  of  arch  ring  13  ft.,  then  the  uniformly  distributed  load  will  be 
1070  pounds  per  square  foot,  and  taking  the  weight  of  one  cubic  foot  of  concrete  equal 
to  ?-  =  140  Ibs.,  then  the  load  in  terms  of  cubic  feet  of  masonry  would  be  1070-:- 140=7.64 
cubic  feet  =  p.  This  includes  67  per  cent  for  impact. 

The  floor,  ballast,  ties,  rails,  etc.,  complete,  will  weigh  710  Ibs.  per  sq.  ft.  according 
to  a  design  previously  made,  and  this  gives  the  dead  load  of  floor  q  =710/140  =5.07 
cu.ft.  of  masonry.  Since  the  floor  contains  considerable  steel,  its  actual  thickness  is 
not  q,  but  in  the  example  chosen  it  was  only  4  ft. 

The  preliminary  design  of  the  arch  ring  is  now  undertaken,  referring  to  Fig.  7lA, 
for  the  lettered  dimensions. 


LIVE   LOAD 


r~-         ^  r 

^T 
^^        d 

-~r.     v>          .r- 

^,         \  1           _,-" 

~~*T 

1 

FIG.  7lA. 

The  thickness  of  the  arch  ring  at  the  crown  is  approximated  from  the  formula?  of 
A.  Tolkmitt,*  which  are  rational  but  approximate,  and,  therefore,  furnish  much  better 
results  than  purely  empiric  formula?.. 

Thus  for  a  uniform  live  load  p/2  over  the  whole  span,  and  an  allowable  unit  stress 
0.4/,  which  is  supposed  to  be  0.4  as  great  as  for  the  unsymmetric  critical  loading,  the 
crown  thickness  should  be 

4+T5)' 

t  0 for  feet  or  meters, (71  A) 


where  the  allowable  unit  stress  /  is  likewise  expressed  in  cu.  ft.  of  masonry,  as  was  done 
for  the  loads. 


*  Leitfaden  fuer  das  Entwerfen  Gewoelbter  Bruecken,  1895. 


ART-  71  FIXED  MASONRY  ARCHES  307 

For  the  unsymmetric  live  load,  extending  over  half  the  span  to  the  center,  the 
following  value  of  Z)0  is  obtained : 


where  G  =  $(q  +  ?-+*°}         f  for  feet  or  meters, (71s) 

\     2i     Au/       j 

When  the  value  from  Eq.  (7 IB)  exceeds  that  given  by  Eq.  (71  A)  then  the  larger 
dimension  should  be  adopted. 

It  is  clearly  seen  that  the  full  allowable  unit  stress  /  could  not  be  taken  in  the 
above  formulae,  because  this  average  case  of  loading  will  stress  the  arch  ring  only  about 
half  as  much  as  when  the  resultant  polygon  passes  through  the  middle  third  points.  In 
all  arch  designs  where  tensile  stresses  are  prohibitive,  this  condition  will  govern. 

The  next  step  is  to  choose  a  preliminary  shape  for  the  intrados  such  that,  for  the 
case  of  dead  load  plus  half  live  load  over  the  whole  span,  the  resultant  polygon  will 
exactly  coincide  with  the  arch  center  line. 

At  this  point  all  theory  fails  and  the  judgment  and  experience  of  the  engineer  must 
guide  in  making  a  suitable  first  approximation.  However,  the  following  practical 
suggestions  will  serve  a  valuable  purpose. 

Theoretically  the  intrados  can  never  be  a  true  circle  nor  a  true  parabola  for  any  arch 
which  is  made  to  follow  the  resultant  polygon,  as  above  required,  for  economic  reasons. 

The  resultant  polygon  becomes  a  parabola  when  the  total  load  is  uniform-  per  foot 
of  arch.  This  can  never  happen  in  a  real  arch,  even  if  the  spandrel  filling  were  neglected, 
because  the  stresses  in  the  arch  ring  increase  from  the  crown  toward  the  abutments, 
thus  necessitating  a  variable  thickness  of  ring,  increasing  with  the  stress. 

On  the  other  hand  the  resultant  polygon  becomes  a  circle  when  the  superimposed 
load  at  the  springing  points  becomes  infinite,  a  condition  which  is  never  attainable. 

Hence  an  arch  of  economic  design  and  shape  to  fit  the  resultant  polygon  for  average 
loading  will  have  an  intrados  which  ^neither  a  circle  nor  a  parabola,  but  a  curve 
lying  between  these  two. 

Having  thus  established  the  limits  between  which  the  true  intrados  must  be  situated 
and  knowing  from  experience  that  the  true  line  falls  nearer  to  the  mean  of  the  two  curves 
than  to  either  one  of  them,  for  all  open  spandrel  arches,  it  becomes  an  easy  matter  to 
approximate  the  intrados. 

Hence,  for  open  spandrels  with  roadway  supported  on  piers  or  columns,  the  intrados 
may  safely  be  taken  half  way  between  the  circle  and  the  parabola,  both  drawn  through 
the  three  given  points  of  the  intrados. 

The  radius  of  the  circle  is  given  by  the  formula 


and  the  parabola  passing  through  the  same  three  points  is 


which  gives  the  coordinates  y  =ho/l  and  x  =/0/4  for  the  quarter  points,  see  Fig.  7U. 


308  KINETIC  THEORY  OF  ENGINEERING  STRUCTURES          CHAP.  XV 

From  this  a  mean  point  at  each  quarter  span  is  readily  determined  and  a  three- 
center  curve  is  then  made  to  pass  between  these  two  and  the  given  crown  and  abutment 
points. 

When  the  spandrel  filling  is  solid,  then  the  intrados  approaches  nearer  to  the  circle 
and  for  a  plain  arch  ring  without  any  spandrel  loading,  the  true  intrados  approaches 
the  parabola.  Between  these  limits  the  designer  will  soon  be  able  to  make  very  close 
approximations  even  for  the  first  trial. 

To  draw  the  extrados,  no  good  rules  can  be  given,  though  a  fair  approximation  can 
be  made  by  applying  the  formula  for  the  thickness  of  any  point  of  an  arch  ring  in  terms 
of  the  thickness  DO  at  the  crown,  as  follows: 

£=— V          /   .-     .      .      .      /»;.;.       (71B) 

cos  9 

For  flat  arches  this  gives  good  approximations,  but  for  semicircular  arches  the 
formula  should  not  be  used  for  angles  greater  than  30°  from  the  crown.  Even  for 
<£=30°  the  values  are  a  little  large  and  for  (/>  =90°,  D  =  GO  ,  which  is  clearly  impossible. 

However,  the  30°  from  the  crown  include  the  most  important  portion  of  the  arch 
ring,  so  that  the  remainder  of  the  extrados  can  be  approximated  with  sufficient  accuracy 
for  the  first  trial. 

Another  method  is  to  sweep  an  arc  with  radius  r'=r+Z)o  through  the  center  of  the 
crown  joint,  which  will  be  a  good  approximation  for  a  center  line  between  crown  and 
quarter  points. 

It  is  not  prudent  to  spend  more  time  in  preliminary  guessing,  but  with  the  dimensions 
above  given  proceed  at  once  to  the  graphic  analysis. 

For  this  purpose  the  arch  should  be  drawn  on  a  scale  of  at  least  1 : 100,  using  a  scale 
1  ft.  =3000  cu.  ft.  of  masonry  for  the  scale  of  forces.  For  small  structures  larger  scales 
may  be  employed. 

The  arch  ring  is  now  divided  up  into  suitable  sections  corresponding  with  the  panel 
lengths  chosen  for  the  spandrel  piers  and  the  areas  of  these  sections  and  of  the  piers  are 
then  computed  from  the  drawing. 

Having  thus  found  all  the  areas,  including  the  uniform  floor  load  q  per  ft.  of  bridge, 
we  may  proceed  to  draw  the  resultant  polygon  for  a  live  load  p/2  extending  over  the 
whole  span.  This  polygon  should  be  passed  through  the  center  points  of  the  crown 
and  abutment  joints  and  if  it  coincides  quite  closely  with  the  assumed  arch  center  line, 
then  the  investigation  may  be  continued  for  the  unsymmetric  loading,  otherwise  the 
shape  of  the  arch  ring  must  be  corrected  and  the  first  resultant  polygon  must  be  recon- 
structed. A  second  trial  nearly  always  gives  acceptable  results,  and  sometimes  the  first 
trial  is  sufficiently  close.  For  symmetric  loading  the  resultant  polygon  is  drawn  for 
the  half  span  only. 

With  the  dead  loads  previously  found  and  a  live  load  covering  half  the  span  and 
extending  past  the  crown  to  the  load  divide  i  for  the  critical  point  m,  see  Fig.  7lA,  now 
pass  a  resultant  polygon  through  the  points  a,  b,  and  n.  The  point  a  is  the  outer  third 
point  of  the  arch  section  at  A,  the  point  b  is  the  inner  third  point  at  B  and  the  point  n 
is  about  Do/8  above  the  center  line  at  the  crown.  This  is  the  first  approximation. 


ART.  72  FIXED  MASONRY  ARCHES  309 

If  this  resultant  polygon  remains  within  the  middle  third  at  every  point  and  crosses 
the  axial  line  in  at  least  three  points,  then  the  design  is  acceptable.  If  the  polygon 
does  not  touch  the  middle  third  point  at  the  two  critical  sections  m  and  s,  Fig.  7lA, 
then  the  ring  is  too  thick.  If  the  polygon  goes  outside  the  middle  third  at  one  of  these 
sections  and  remains  inside  at  the  other,  then  the  point  n  should  be  shifted  and  a  new 
polygon  should  be  passed  through  a,  b,  and  n.  However,  if  the  resultant  polygon 
passes  outside  the  middle  third  at  both  critical  sections,  then  the  arch  ring  must  be 
made  thicker  and  the  investigation  is  then  repeated. 

Lastly  the  stresses  at  the  four  sections  a,  m,  n,  and  b  must  not  exceed  the  allowable 
unit  stress  as  given  by  Eq.  (69A). 

When  all  these  requirements  are  fulfilled,  the  crown  section  should  be  tested  for 
full  live  load  over  the  entire  span,  which  is  easily  done  from  the  loads  already  found  for 
the  half  span,  since  this  is  a  case  of  symmetric  loading. 

This  then  constitutes  a  complete  design  according  to  the  ordinary  graphic  method, 
and  the  arch  ring  so  found  should  now  be  subjected  to  a  rigid  analysis  according  to  the 
theory  of  elasticity  which  follows. 

The  above  outline  will  be  exemplified  more  fully  when  presenting  a  problem  at  the 
close  of  this  chapter. 

The  criterion  for  position  of  moving  loads  to  produce  maximum  and  minimum  stress 
is  discussed  in  another  <place. 

ART.  72. 

Introductory.  A  fixed,  solid  web  arch  must  be  treated  as  a  structure  with  three 
external  redundant  conditions  according  to  Eq.  (3c)  and  Fig.  SF.  Hence  there  is  very 
little  difference  between  a  fixed  framed  arch  and  a  fixed  solid  web  or  masonry  arch, 
since  the  complications  arising  from  external  redundancy  are  alike  in  both  cases  while 
the  internal  stresses  must  be  found  by  the  methods  peculiar  to  frames  and  isotropic 
bodies,  respectively. 

The  method  of  treating  the  external  redundant  conditions  involves  the  use  of 
deflection  polygons  for  the  determinate  principal  system,  and  in  this  connection  a 
further  difference  exists  in  the  manner  of  determining  the  elastic  loads  for  the  two  types 
of  fixed  arches. 

The  elastic  loads  w  for  a  framed  structure  are  easily  expressed  in  terms  of  the 
geometric  relations  existing  between  the  members  and  the  angles  composing  the  frame. 
For  isotropic  bodies  the  elastic  loads  are  functions  of  the  cross-sections  obtained  by 
integration  and  expressed  in  terms  of  moments. 

Hence,  with  due  regard  to  these  differences,  \vhich  are  easily  distinguished,  both 
framed  and  solid  web,  fixed  arches  may  be  analyzed  in  precisely  the  same  manner. 
However,  the  general  subject  of  solid  web  and  masonry  arches  involves  many  consider- 
ations not  met  with  in  framed  arches.  For  this  reason  the  solid  web  arch  will  be  treated 
in  full  detail  in  the  following  articles. 

Regarding  the  reliability  of  the  method,  which  is  based  on  the  theory  of  elasticity, 


310 


KINETIC  THEORY  OF  ENGINEERING  STRUCTURES 


CHAP.  XV 


the  author  would  add  here  that  for  steel  arches,  either  framed  or  solid  web,  the  method 
is  entitled  to  full  confidence,  while  for  stone  and  concrete  it  is  reliable  when  the  design 
is  based  on  the  condition  that  tensile  stress  does  not  occur.  In  attempting  to  design 
in  concrete  for  equal  tensile  and  compressive  stress  by  reinforcing  with  steel,  the  method 
must  be  regarded  as  more  or  less  approximate,  since  it  involves  the  modulus  E  of  the 
material,  which  is  quite  uncertain  owing  to  the  heterogenous  character  of  masonry  and 
concrete  generally,  and  especially  when  the  latter  is  combined  with  steel.  The  intro- 
ductory remarks  of  Art.  52  are  especially  applicable  here. 

General  relations  of  the  external  forces  to  the  principal  system.  The  considerations 
under  this  heading,  in  Art.  52,  are  again  applied  to  the  analysis  of  the  solid  web  arch, 
and  the  principal  system  is  chosen  as  a  determinate  beam  on  two  supports. 


FIG.  72A. 

Fig.  72A  shows  the  principal  system  for  an  unsymmetric  arch  with  the  redundant 
conditions  acting  as  external  forces  on  the  left-hand  abutment.  Similar  redundants 
act  on  the  right-hand  abutment,  and  these  are  equal  and  opposite  to  the  set  shown  for 
the  left-hand  end.  See  Figs.  52n  and  52c. 

The  arch  ring  aa'bb',  loaded  with  a  single  load  P,  is  referred  to  coordinate  axes 
(x,  y)  choosing  the  y  axis  vertical  and  the  x  axis  making  some  angle  /9  with  the  horizontal. 
The  origin  0  is  supposed  to  be  known  and  is  fixed  by  certain  geometric  relations  to  be 
established  later. 

The  ordinate  ym,  of  any  point  ra  of  the  gravity  axis  or  axial  line  of  the  arch  ring, 
is  measured  vertically  Irom  the  x  axis,  while  the  abscissa  xm>  of  this  point,  is  measured 
horizontally  from  the  y  axis  instead  of  parallel  to  the  x  axis.  This  is  a  mere  matter  of 
convenience. 

The  span  AB=l  is  taken  as  the  line  joining  the  intersections  between  the  axial  line 


ART.  72  FIXED  MASONRY  ARCHES  311 

and  the  verticals  through  a'\  and  b'\.  The  support  at  a\  is  assumed  as  hinged  while 
the  one  at  b\  is  made  movable  for  the  principal  system. 

The  single  load  P  then  produces  reactions  RI  and  RZ,  intersecting  in  the  point  C 
and  passing  respectively  through  the'  unknown  points  a^  and  61  on  the  verticals  through 
the  supports  a\  and  b\.  The  triangle  aiC&i  thus  becomes  a  resultant  polygon  with 
the  closing  line  a\b\.  The  reaction  RI  may  be  resolved  into  the  vertical  component  A0 
and  the  haunch  thrust  H'  along  a\b\.  The  reaction  R%  may  be  similarly  resolved  into 
the  vertical  reaction  B0  and  the  thrust  H',  which  latter  is  equal  and  opposite  to  H' 
acting  at  a\. 

The  vertical  reactions  A0  and  B0  are  the  same  as  for  a  simple  beam  of  span  I  on 
determinate  supports.  Hence, 

A0=j(l-e)       and  '  B0=*£.     .     .......     (72A) 

Also,  the  moment  for  any  point  m  of  the  simple  beam,  equals 

-\f  i\f 

osa       or       tf--i--,    ....     (72B) 


where  K  is  the  ordinate  ig  through  m,  and  H  is  the  horizontal  component  of  H'  . 

Therefore,  the  resultant  polygon  a\Cb\  becomes  fixed  whenever  H'  and  the  closing 
line  a\b\  are  found. 

The  origin  0  is  connected  to  the  arch  along  aa'  by  the  rigid  disc  aa'O  and  to  this 
origin  are  applied  two  equal  and  opposite  forces  Hf,  which  are  equal  and  parallel  to  the 
original  hanuch  thrust  acting  at  a\.  The  equilibrium  of  the  principal  system  and  of 
the  fixed  arch  thus  remains  undisturbed. 

Suppose  now  that  all  the  external  forces  to  the  left  of  a  section  tt'  act  on  the 
principal  system  only,  and  that  the  three  forces  H'  and  the  vertical  reaction  A0  are 
applied  to  the  rigid  disc  and  are  thence  transmitted  to  the  principal  system.  Then  the 
force  H'  at  a]  and  the  opposing  force  H'  acting  at  0,  form  a  couple  with  lever  arm 
z0cosa,  producing  a  moment  Xa  =H'z0cosa.  The  other  force  H',  acting  at  0  and  to 
the  right,  may  be  resolved  into  two  components  Xb  and  Xc,  where  Xb  is  vertical  along 
the  y  axis,  and  Xc  acts  along  the  x  axis.  The  external  forces  to  the  left  of  the  section 
and  applied  to  the  principal  system  are  then  P,  A0,  Xb,  Xc,  and  a  moment 
Xa=H'z0  cos  a  =  Hz0.  Of  these  the  two  forces  Xb  and  Xc  and  the  moment  Xa  constitute 
the  redundant  conditions  while  the  forces  P  and  A0  are  known,  and  all  are  applied  to 
the  principal  system  to  the  left  of  the  section  tt'.  A  similar  set  of  external  forces  (not 
shown)  acts  on  the  principal  system  to  the  right  of  this  section. 

The  moment  of  all  the  external  forces  abou'c  any  point  m  of  any  structure,  involving 
three  external  redundant  conditions,  is  given  by  Eq\  (7  A)  as 

Mm^Mom-MaXa-MbXb-McXc,     .......     (72c) 

wherein 

Mom=-A0(li  —  xm)—Pd=the   moment   about   m  due   to   the   load   P  acting    on   the 
principal  system,     This  is  condition  X=0. 


312  KINETIC  THEORY  OF  ENGINEERING  STRUCTURES          CHAP.  XV 

Ma  =  l  is  the  moment  about  m  due  to  the  moment  Xu  =  l  applied  to  the  principal 

system.     Condition  Xa  —  1 . 
Mb  =  1  •  xm  is  the  moment  about  m  due  to  the  force  Xb  =  1  acting  on  the  principal 

system.     Condition  Xb  =  1 . 
Mc  =  l-ym  cos  /?  is  the  moment  about  m  due  to  the  force  Xc  =  l  acting  on  the 

principal  system.     Condition  'Xc  =  1 . 

Substituting  these  values  into  Eq.  (72c),  gives  the  following  fundamental  moment 
equation  for  fixed  arches: 

Mm=Mom-l-Xa-xmXb-ymcospXe ,     .     (72D) 

Before  Mm  can  be  determined  for  any  point  m  of  the  arch,  the  three  redundants 
Xa,  Xb,  and  Xc  must  be  evaluated  from  three  simultaneous  work  equations  of  the  form 
of  Eqs.  (44s)  which  may  be  made  to  apply  to  the  present  problem  by  so  locating  the 
(x,  y)  axes  that  dab=dba=Q,  ^=^=0  and  dbc=dcb=0. 

The  Redundant  conditions  for  a  single  load  P  —  1  then  become 

^w>f  "  "w»Z> "  "me 


for  fixed  abutments  and  omitting  temperature  effects. 

Neglecting  the  term  involving  the  axial  thrust  N  in  Eq.  (15N)  the  deflections  in 
Eqs.  (72E)  may  be  expressed  as  virtual  work  in  terms  of  the  moments  Mom,  Ma  =  l, 
Mb=x  and  Me=y  cos/?,  as  follows: 

~Momdu  C  Mldu      Cdu 


_  fMomMadu  _  ( 
dma~J~~~ET    ~J 

rMomMbdu      fMomxdu  C 

8«*  =)  -  -Er   -J  -^7—;  fa  - J 


El 

j~.  r  ~\/r1j,.        r-.vj,. 

(72r) 


El 

CMomMcdu      CMomy  cos  8du  C  M2cdu      A/2  cos2  3du 

Omc=J ^J—  Ej          -I  ^=J   — EJ~   =J gf- 

Introducing  the  elastic  loads 

du  du  du 

v7=wa,      x~T^T=wb,       and       y-j^f  =wc (72c) 

Hit  til  '  hi 

into  Eqs.  (72r),  then  substituting  these  values  into  Eqs.  (72E)  and  replacing  the  inte- 
grations by  summations,  then 


V   _m    _          -  omb  -^    _     mc_      jomc  .          . 

~~~  ;       Ac~~ 


The  modulus  E,  being  involved  in  both  numerator  and  denominator,  of  Eqs.  (72n), 
has  no  effect  on  the  redundants  except  when  temperature  and  reaction  displacements 
are  included. 

If  the  axial  thrust  is  to  be  considered,  then  the  deflections  due  to  this  effect  alone 


ART.  72 


FIXED  MASONRY  ARCHES 


313 


and   for   Na=Q,   A\  =  1  •  sin  (f>   and   A'c  =  1  •  cos  </>,    according   to   Eq.    (15N),   become   for 
du=dx/cos  (f): 


/AT~   /  fl        '      2  rk  A 

-fij-  =  J0  Ep  CO^A 

fNc2du  _  fcos2  fi  du  _  Cl  cos  <£  Ax 
CC=J  ~Ejr=J       EY        Jo  ~EF~ 


(72j) 


These  values  added  for  total  effect  includin     axial  thrust  in  the   determination  of 


the  redundants  will  change  Eqs.  (72n)  to  the  following: 

Xa=- 


I,Momwc 


sn 


(72K) 


EF  cos  (f> 


Eqs.  (72n)  and  (72x)  being  written  for  a  single  load  P  =  l  represent  the  equations 
of  the  influence  lines  of  the  redundants. 

If  axial  thrust  is  to  be  considered  in  Eqs.  (72K)  it  will  be  well  to  compute  the  two 
functions  in  the  denominators  and  thus  include  them  in  the  pole  distances  when  drawing 
the  Xb  and  the  Xc  influence  lines. 

It  should  be  pointed  out  that  in  general,  axial  thrust  becomes  important  for  very 
flat  arches,  in  which  case  the  function  S  sin2  <f)Ax/EF  cos  $  becomes  very  small,  while 
£  cos  <f>Jx/EF  becomes  large.  Hence  it  is  never  necessary  to  include  the  thrust  function 
in  determining  X^,  while  for  Xc  it  may  take  on  considerable  proportions. 

Regarding  the  curvature  of  an  arch,  which  is  involved  in  the  integral  J  du  and  which 

could  be  fully  considered  should  the  necessity  arise,  it  must  be  remembered  that  the 
depth  of  the  arch  ring  is.  almost  without  exception,  very  small  compared  with  the 
radius  of  curvature  of  the  axial  line,  and  no  appreciable  error  is  committed  when  this 
effect  is  entirely  neglected.  Some  comparative  analyses  have  been  worked  out  by 
various  investigators,  which  show  quite  conclusively  that  this  error  is  smaller  than  the 
knowable  accuracy  of  the  strength  of  materials,  to  say  nothing  about  temperature 
stresses  and  abutment  displacements,  which  may  be  estimated  but  never  become  known. 

For  these  reasons,  which  are  considered  ample,  the  question  of  axial  curvature  will 
not  be  treated  here. 

The  location  of  the  coordinate  axes  was  to  be  so  chosen  that  the  displacements 
o<j6  =  ^ar"=^c;==0,  thus  permitting  the  use  of  the  simplified  Eqs.  (44s)  instead  of  the 
involved  form  of  Eqs.  (8b) . 

According  to  Eq.  (15isT)  and  noting  that  Na=0,  these  displacements  become 


fMaMbdu      Cxdu      v 

ab==J  — TH~  =J  ~w  =^x' 

CMaMcdu_  (ydu 
dac~J       El      "JET      ^y 

CMbMcdu      fxydu     v 
bc  =J  — ~KL~    =  J    FT  =  ^x 


,o   =2wc=0 
w«=0 


314 


KINETIC  THEORY  OF  ENGINEERING  STRUCTURES 


CHAP.  XV 


If  the  coordinate  axes  are  located  so  as  to  comply  with  the  conditions  of  Eqs.  (72L) 
then  the  origin  of  these  axes  must  be  the  center  of  gravity  of  the  elastic  loads  wa  because 
%xwa=Q  and  '£i/wa=Q.  Also  the  axes  must  be  conjugate  in  order  that  the  centrifugal 
moment  ^lxywa—0. 

The  manner  of  computing  the  elastic  loads  and  then  of  finding  the  positions  of  the 
axes  with  respect  to  any  unsymmetric  arch  ring  follows  here. 

ART.  73.     COORDINATE   AXES   AND   ELASTIC   LOADS 

Unsymmetric  arches.  The  elastic  loads  wa  depend  on  the  geometric  shape  of  the 
arch  sections  while  the  loads  Wk—xwa  and  wc=ywa  are  seen  to  be  functions  of  the  wa 
loads  and  of  the  coordinates  of  the  axial  points  of  the  arch.  Hence  the  wa  loads  may 
be  found  for  any  arch  ring  of  given  sections  while  the  w^  and  wc  loads  require  for  their 
determination  a  careful  location  of  the  coordinate  axes  in  compliance  with  the  conditions 
given  by  Eqs.  (72b) . 

The  first  step  is  then  to  evaluate  the  wa  loads  for  finite  distances  Ju,  which  is  done 

by  dividing  up  the   arch  ring    into    an 
even  number  of  lengths  and 
puting   El   for  each  section, 

Wa  ••=  I  wa  =  I      -=-7  are  found  by  summa- 
J          Jo    Jiil 

tion  according  to  Simpson's  rule.  This  will 
answer  all  practical  purposes  and  will  be 
the  more  accurate  the  shorter  the  lengths 
Au  or  the  greater  the  number  of  sections 
taken. 

The  manner  in  which  this  is    done 
will  now  be  illustrated,  as  the  accuracy 
of  the  analysis  depends  almost  entirely  on  the  values  Wa.     See  Fig.  73 A. 

Instead  of  dividing  the  axial  line  into  small  lengths  Ju,  it  is  preferable  to  make 
these  spaces  equal  horizontally,  so  that  Ju  =  Ax/cos  (f>,  where  <£  is  the  angle  which  the 
normal  section  makes  with  the  vertical.  The  Eqs.  (72c)  then  become 

Au          Ax  xAx  ,  yAx 


after  com- 
the  values 


Ix'^ix* 


T 
I 

X»        ,   ^ 

TN    * 

•  M 

i  '   »  i  - 

I't-M 

'I 

i 

If 

B 
+\ 

*•** 

i 

x^x^x-*.x4 

»•*»-" 

i  - 

FIG.  73A. 


Wa    El     El 


wb  =xwa  = 


El  cos 


and 


El  cos 


—> [•     •     (73  A) 


Each  half  of  the  arch  is  now  spaced  off  into  an  even  number  of  spaces  Ax,  beginning 
at  the  crown.  When  the  spaces  Au  appear  to  get  too  large  they  may  be  shortened 
toward  the  haunches  of  the  arch,  but  it  is  best  to  keep  as  many  as  possible  of  the  Ax  alike. 

Now  tabulate  the  depths  D  of  the  arch  ring  and  angles  <£  for  the  axial  points  from 
A  to  B  and  figure  the  reciprocal  values  of  El  cos  <j>  for  each.  If  the  section  is  one  of 
a  steel  girder  the  real  moment  of  inertia  is  to  be  used  while  for  a  masonry  arch  it  is  best 
to  take  a  section  of  width  unity  and  depth  D  making  I=--D3/12. 

Simpson's  rule  is  now  employed  to  sum  the  values  wa  in  order  to  obtain   finite 

r/*    I  x»  / 

w===~~~~ for  the 


ART.  73 


FIXED  MASONRY  ARCHES 


315 


The  quantities  1/EIcoscf)  are  treated  as  parallel  ordinates  to  some  irregular  curve 
and  the  volume  of  length  I,  which  is  divided  into  an  even  number  n  of  equal  spaces  Ax-, 
is  then  iven  from 


Wa=-^-[w0+4wi+2w2+4:W3+2w4  to  4u'n_i 

o 


(73s) 


One  such  compression  can  be  written  for  any  even  number  of  equal  spaces  Ax,  Ax', 
or  Ax". 

To  find  Wa=\wa  for  each  of  the  sections,  Simpson's  rule  for  three  quantities  is 
then  applied.  This  is  the  well-known  prismoidal  formula  for  volume  and  is  used  here 
to  figure  volumes  between  successive  mean  areas  taking  the  originally  computed  values 
w  for  the  middle  terms. 

Since  the  loads  Wa  are  wanted  for  the  points  m  themselves  and  not  between  these, 
we  interpolate  mean  values  between  consecutive  pairs  of  computed  values  and  then 
figure  the  sums  Wa,  using  the  computed  Wa  as  the  middle  term  in  the  formula. 

Thus  the  formula, 

Ax 
wa3=-^-[w2+4:W3+w4], (73c) 

as  here  applied,  becomes 

Ax  \w2  +ws  ,  w3  +WA~\ 

w^=-^-    -V--+4w3+-         M, (730) 

6  [      2  2      J 

when  the  interpolated  means  are  used.     This  gives  a  much  better  check  on  Eq.  (73s) 
than  could  be  obtained  by  the  use  of  Eq.  (73c). 

Beginning  at  A,  and  making  —  - — I.=w0_i,  -      — -=wi_2,  etc.,  the  several  values 

£  £ 

of  Wa  are  as  follows: 

Ax' 
i'W«o  =—^-\2wo  +WQ _il  =  half  load  for  an  end  section 

D 
W      -  — 

6 


Wa2  =  -         _ 


2 


Ax 

a3  =—  [MJ2_3 


Ax 


etc., 

w        ^x 
Was=-Q- 

etc., 


etc. 

Ax" 

•_8  +2lV8]  +——[2Wg  +W>8_9 

etc. 


=- -Wn  -12 


(73E) 


316  KINETIC  THEORY  OF  ENGINEERING  STRUCTURES 

and  as  a  final  check  the  sum  of  all  Wa  loads  must  be  by  Eq.  (73s) 


CHAP.  XV 


Xi         AX'  jx 

wa  =-Hi00  +±wi  +w2]  +  -x- 
o  o 


4  +4w5  +2w6 

Ax" 
+4w7  +w8]  +—  —  [ws  +4wg  +2ww  +4wn 

o 


/dx  CAx 

wa.  these   are   now  used 
El  cos  0     Jo 

to  determine  the  center  of  gravity  0  and  the  angle  /?  which  the  x  axis  makes  with  the 
horizontal.  The  y  axis  was  taken  vertically  and  is  thus  fixed  as  soon  as  the  origin  is 
located. 


+v 


FIG.  73s. 


The  center  of  gravity  0  is  located  with  respect  to  any  assumed  rectangular  axes 
(z,  v),  Fig.  73B,  with  origin  at  the  axial  point  A  the  same  as  shown  in  Fig.  73A.  It  is 
most  convenient  to  assume  the  z  axis  vertical  and  the  v  axis  horizontal.  The  coordinates 
(z,  v)  are  found  for  each  of  the  several  axial  points  m  previously  used  for  the  computation 
of  the  Wa  loads,  and  all  are  tabulated  together.  Finally  the  moments  zWa  and  vWa  are 
found  and  from  these  the  coordinates  l\  and  z,7  of  the  origin  0  are  obtained  from 


— /  v  rr  a  ,  .         *-*a 

ZI=^TI^;    and    «• -^ 


(73F) 


This  also  fixes  the  y  axis  which  is  parallel  to  the  vertical  z  axis  through  the  center  of 
gravity  0.  The  x  axis,  while  passing  through  0,  makes  some  angle  /?  with  the  horizontal 
such  that  I,xyWa=Q,  according  to  the  last  of  Eqs.  (72L). 

The  (x,  y)  coordinates  may  be  derived  from  the  (z,  v)  coordinates  when  the  angle 
,5  is  determined.  Taking  /?  positive  when  measured  to  the  right  of  the  origin  and  above 
the  horizontal,  as  shown  in  Fig.  73s,  then 


x  =li  —v  and  y  =z  —z0'  +x  tan  /? 


(73o) 


ART.  74 


FIXED  MASONRY  ARCHES 


317 


The  angle  /?  is  found  by  substituting  the  value  of  y  from  Eq.  (73c)  into  the  condition 
equation  %xyWa=Q,  giving,  as  in  Eq.  (52j), 


(73H) 


The  abscissae  x  being  known  from  Eqs.  (73c)  the  values  HxzWa  and  2x2Wa  are 
readily  found  and  tan  /?  is  then  obtained  from  Eq.  (73n)  and  the  new  axes  and  (x,  y) 
coordinates  are  thus  determined. 

For  symmetric  arches  the  above  calculations  are  greatly  simplified  and  the  coordinate 
axes  are  readily  located  as  follows:  The  y  axis,  being  the  axis  of  symmetry,  is  practically 
given  when  the  shape  of  the  arch  is  given,  and  the  angle  0=0  because  for  symmetric 
forces  the  centrifugal  moment  (  xyWa  =0  for  any  pair  of  rectangular  gravity  axes. 

Hence  the  y  axis  and  the  direction  of  the  x  axis  are  given  and  the  origin  0  is  located 
by  computing  the  ordinate  z0'  from  any  assumed  horizontal  v  axis,  giving  for  symmetric 

arches, 

SzW 
'  ;       0=0;       and       y=z-z0r  .......     (73j) 


In  any  case  the  Wa,  Wb,  and  Wc  loads  are  now  readily  figured  and  from  these  the 
equations  for  the  influence  lines  for  the  redundant  conditions  Xa,  Xb,  and  Xc  are  evaulated. 

Solid  web  circular  arches  which  are  only  very  slightly  unsymmetric  may  be  approx- 
imately analyzed  by  extending  the  short  half  to  a  point  of  symmetry  and  treating  the 
structure  as  a  symmetric  arch.  However,  when  the  difference  in  the  elevations  of  the 
springing  points  is  appreciable,  such  approximations  should  not  be  made. 

The  center  of  gravity  0  may  be  found  graphically  by  combining  the  Wa  loads  into 
two  equilibrium  polygons,  for  the  loads  acting  first  vertically  and  then  horizontally 
through  the  centers  of  gravity  of  the  arch  sections.  The  two  resultants  thus  obtained 
will  intersect  in  the  required  point  0.  The  moments  of  inertia  5ix2Wa  and  Hiy2Wa  may 
be  obtained  from  the  same  polygons  by  Professor  Mohr's  inethod. 


ART.  74.     INFLUENCE   LINES   FOR   Xa,   X6,   Xc,   AND   Mm 

Having  located  the  coordinate  axes  (x,  y)  to  fulfill  the  requirements  dab—^ae  —  ^bc= 
the  Eqs.  (72E)  and  (72H)  now  become  applicable  to  the  present  problem,  giving 


Xa  = 


~"2W  ~JT~riar 

£j  vv  &        J-^a 

1<\  <N 

U mo        (-*  mo 

~  ~^rW,  =~tf7  =  ^6"; 


cos 


. f\ 

—  -Jj          '/cm 


(74A) 


These  Eqs.  (74A)  which  are  written  for  a  load  Pm  =  l,  thus  represent  the  equations 
for  the  influence  lines  of  the  redundants  Xa,  Xb,  and  Xc. 


318  KINETIC  THEORY  OF  ENGINEERING  STRUCTURES          CHAP.  XV 


For  Xa  =  l,  the  first  of  Eqs.  (74A)  gives  l-^Wa  =  l'dma  =  ^MomWa.  Therefore, 
the  special  moment  Mom  equals  unity  and  this  equals  the  moment  Xa  =  1  acting  on  the 
principal  system,  which  would  follow  from  Maxwell's  law.  Hence,  an  equilibrium 
polygon  drawn  for  the  loads  Wa  with  pole  unity,  would  give  the  deflection  polygon 
for  dma  due  to  a  moment  Xa  =  I  applied  to  the  principal  system.  Also,  the  ordinates 
of  a  similar  equilibrium  polygon  drawn  with  a  pole  Ha  =  ^Wa  would  represent  the 
ordinates  rjaw,  of  the  Xa  influence  line,  measured  to  the  scale  of  lengths  of  the  drawing. 

The  influence  lines  for  Xj,  and  Xc  are  similarly  found  from  equilibrium  polygons 
drawn  for  the  elastic  loads  Wb  and  W&  with  pole  distances  respectively  equal  to  2xWb=Hb 
and  cos  pEyWc=Hc.  When  axial  thrust  is  to  be  considered  then  the  pole  distance 
Hc=cos  p^yWc  +  S  cos  2<j>du/EF  =cos  ^yWc  +  2  cos  <j>Ax/EF  from  Eq.  (72K)  . 

The  Mm  influen.ce  line  for  moments,  about  any  point  m-  of  the  axial  line  and  a 
moving  load  Pm=»l,  is  now  derived  from  Eq.  (72D).  Accordingly  the  ordinates  jjm, 
of  the  Mm  influen.ce  line,  must  represent  the  algebraic  summation  of  the  influences  due 
to  Mom,  Xa,  Xb,  and  Xc  expressed  by  the  equation 

ym=Mm=Mom-Xa-xmXb-ym  cos  t8Xc    ......     (74s) 

The  influence  line  for  Mom  is  the  ordinary  moment  influence  line  drawn  for  the 
point  m  and  for  a  simple  beam  on  two  determinate  supports  at  a'\  and  b'i,  Fig.  72A. 
It  is  a  different  line  for  each  axial  point  m. 

The  influence  lines  for  Xn,  Xb,  and  Xc  remain  the  same  for  the  same  structure,  but 
the  ordinates  of  the  Xb  and  Xc  lines,  according  to  Eq.  (74s),  require  multiplication  by 
the  variable  coordinates  x  and  y  cos  {3,  which  differ  for  each  axial  point  m. 

Hence  the  Mm  influence  line  is  best  constructed  by  computing  the  ordinates 
(Xa+xmXb+ym  cos  3XC)  for  the-  particular  values  of  xm  and  ym  cos/?  and  plotting  these 
ordinates  negatively  from  the  Mom  influence  line. 

The  Mm  line,  so  found,  will  serve  to  determine  the  bending  moment  for  the  axial 
point  m  due  to  any  position  of  a  system  of  concentrated  loads. 

The  normal  thrust  Nm,  and  stresses  on  any  normal  arch  section  through  m,  are 
then  computed  as  described  in  Art.  49  and  further  discussed  below  by  involving  the 
moments  about  the  kernel  points  e  and  i. 


ART.  75.     TEMPERATURE   STRESSES 

The  effect  ,of  a  uniform  rise  in  temperature  is  to  expand  the  principal  system 
uniformly  in  all  directions.  But  since  a  uniform  vertical  expansion  is  not  resisted  by 
the  fixed  abutments,  it  may  be  assumed  that  the  whole  temperature  stress  is  produced 
by  horizontal  expansion.  This  manifests  itself  by  a  bending  moment  extending  over 
the  entire  arch  ring,  and  produced  by  resisting  abutments. 

Thus  for  a  rise  in  temperature,  the  extreme  elements  of  the  extrados  at  the  crown 
are  in  tension  while  at  the  haunches  they  are  in  compression,  when  the  arch  is  not 
otherwise  loaded.  The  simultaneous  condition  of  the  intrados  elements  is  exactly  the 
reverse. 


ART.  75  FIXED  MASONRY  ARCHES  319 

In  Eqs.  (72E)  the  temperature  term  was  neglected.  It  is  now  separately  considered 
without  regard  to  the  effect  of  the  loads  P. 

According  to  Eqs.  (44c),  which  apply  to  the  present  problem,  the  redundant 
temperature  stresses,  for  rigid  abutments,  become: 

Xat=^=0;      A"M=A'=0;      and      Xct=d^  .......     (75A) 

Oaa  Obb  Occ 

The  first  two  are  placed  equal  to  zero  b.ecause  uniform  temperature  changes  cannot 
produce  rotation  nor  -vertical  deflection  of  the  principal  system,  which  latter  is  here 
regarded  as  a  simple  beam  on  two  supports. 

The  distance  over  which  the  change  dct  is  cumulative  must  be  the  projection  of  the 
span  /  on  the  x  axis,  hence  dct  =etl/cos^  and  from  Eqs.  (72F)  dcc  =  2?/  cos  2(3WC,  giving 
for  the  last  of  Eqs.  (75A)  : 

'  where    w--*w'-  •  •  •  (75B) 


The  moment  about  any  axial  point  m  due  to  temperature  alone,  may  now  be  found 
from  Eq.  (72o),  wherein  Mow=0,  Xat^0,  Xbt=Q,  and  X^  is  given  from  Eq.  (75s). 
Then 

-Mmt  =ym  cos  ?Xet  =^2  *.        .     .     .    ,   ..     .     (75c) 


According  t'o  Eq.  (75c)  the  only  variable  affecting  the  moment  Mmt  is  the  ordinate 
ym.  Hence  for  ym  =0,  Mmt  =0  and  for  ym  maximum,  Mmt  becomes  maximum.  The 
points  where  the  x  axis  intersects  the  axial  line  are  thus  points  of  zero  moment  for 
temperature  effects,  while  the  crown  and  haunch  points  receive  maximum  effects. 

The  temperature  effects  must  finally  be  combined  with  the  dead  and  live  load 
stresses  to  obtain  absolute  maxima  and  minima. 

It  is  a  well-established  fact  that  when  masonry  arches  open  up  cracks  this  generally 
occurs  during  the  coldest  winter  weather.  This  is  because  masonry  or  concrete  is 
scarcely  ever  placed  during  freezing  temperatures,  but  generally  in  the  warmest  summer 
months.  Hence,  the  variations  in  temperature  from  that  existing  at  the  time  of  closing 
the  arch  ring  are  nearly  always  greater  in  the  negative  direction.  Thus,  if  the  ring  was 
closed  at  70°  F.  the  completed  structure  might  at  some  future  period  attain  a  temperature 
of  perhaps  90°,  but  it  is  quite  certain  to  fall  to  about  —10°  if  the  climate  is  severe. 
This  then  would  subject  the  arch  to  temperature  stresses  resulting  from  +20°  and  —80° 
from  the  closing  temperature. 

When  it  is  considered  that  a  lowering  in  temperature  will  induce  tensile  stresses 
in  the  spandrels  and  thus  cause  rupture  at  the  points  of  least  strength,  it  is  clear  why 
this  phenomenon  is  of  such  frequent  occurrence.  Rising  temperature  would  tend  to 
increase  the  compressive  stresses,  but  probably  not  to  any  alarming  extent. 


320  KINETIC  THEORY  OF  ENGINEERING  STRUCTURES          CHAP.  XV 


ART.  76.     STRESSES   ON   ANY   NORMAL   ARCH   SECTION 

This  subject  is  fully  discussed  in  Art.  49  and  illustrated  by  Fig.  49s  and  will  not 
be  repeated  here  except  in  so  far  as  to  show  the  application  of  Eqs.  (49s)  to  the  fixed 
masonry  arch. 

For  any  rectangular  arch  section  the  kernel  points  e  and  i  are  located  on  opposite 
sides  of  the  axial  line  at  equal  distances  k=D/Q  from  the  axis.  The  kernel  point  i  for 
the  intrados  is  situated  above  the  gravity  axis,  while  the  kernel  point  c  for  the  extrados 
is  below  this  axis  as  shown  in  Fig.  49s. 

The  moments  Me  and  Mt-  are  found  from  Eq.  (74s)  in  which  the  coordinates  of  the 
axial  point  m  are  replaced  by  the  coordinates  (xe,  ye)  for  an  extrados  kernel  point  and  by 
the  coordinates  (xi,  yj)  for  an  intrados  kernel  point.  This  is  permissible  because  this 
equation  is  perfectly  general  and  is  true  for  any  point  in  the  plane  of  the  structure. 

The  moment  equations  for  the  kernel  points  e  and  i,  of  any  normal  section,  are 
thus 

Me=Moe-[Xa+xeXb+yec.oat3Xe]    j  ( 

Mi  =  Moi  -[Xa  +  XiXb  +  yi  cos  pXe]    I 

Designating  all  influence  line  ordinates  by  y  and  using  proper  subscripts  as  a 
distinctive  feature,  then  the  ordinates  of  the  kernel  point  moment  influence  lines  may 
be  expressed  as 


wherein  )j0,  x,  and  y  are  special  for  each  normal  arch  section  it'  while  the  ordinates  jja, 
jjfe,  and  rjc  are  the  same  for  all  influence  lines  but  differ  for  each  point  of  the  span. 

When  the  kernel  moments  are  known  for  any  rectangular  section  and  due  to  any 
position  of  a  moving  train  of  loads,  the  stresses  on  the  extreme  fibers  of  the  arch  section 
and  the  normal  thrust  N  and  its  distance  v  from  the  axial  line,  may  be  determined 
from  Eqs.  (49s)  thus: 


J)2    ' 


(76c) 


i=5;      F^l-D;      and      /=~ 

where  D  is  the  depth  and  F  is  the  area  of  a  normal  arch  section  of  unit  thickness. 


ART.  76 


FIXED  MASONRY  ARCHES 


321 


In  Eqs.  (76c)  M0  and  Mi  have  opposite  signs  when  TV  acts  between  the  two  kernel 
points  e  and  i.  The  stresses  fe  and  /;  take  their  signs  from  Me  and  Mi  respectively, 
and  compression  is  regarded  as  a  negative  stress.  The  offset  v  is  positive  when  measured 
from  the  gravity  axis  toward  the  extrados  and  when  so  applied  determines  a  point  on 
the  resultant  polygon  for  that  section. 

It  is  not  necessary  to  draw  the  M,n  influence  lines  because  the  M-e  and  Mi  Tines 
cannot  be  derived  from  the  Mm  line,  nor  is  it  possible  to  determine  N  and  v  when  Mm 
alone  is  known  except  by  constructing  a  resultant  polygon. 

Hence  the  stresses  on  any  arch  section  are  best  found  from  the  kernel  point  moments 
Me  and  Mi  of  that  section.  If  the  redundants  Xa,  X^,  and  Xc  are  evaluated  from  their 
respective  influence  lines  for  a  certain  position  of  a  train  of  loads,  then  the  kernel 
moments,  for  any  section,  may  be  computed  from  Eqs.  (76A).  However,  it  is  preferable 
to  conduct  the  entire  solution  by  means  of  influence  lines,  as  will  be  illustrated 
later. 

Since  different  portions  of  the  arch  ring  are  situated  in  the  four  quadrants  of  the 
coordinate  axes,  it  may  be  well  to  show  how  to  pass  from  the  coordinates  of  any  axial 
point  m  to  those  of  the  kernel  points  of  the  normal  section  through  ra. 

For  a  rectangular  arch  section  ke=ki  —  D/Q,  so  that  the  following  relations  may 
be  written  out,  from  Fig.  76A,  in  terms  of  the  lettered  variable  dimensions  D  and  <f> 
for  each  point,  thus: 


For  point  A,         xe=     x—-s' 


For  point  m, 


For  point  nj 


xe  =     x  —  -~-  sin 

Xi  =     x+-^-sm 
6 


xe=0 


and 
and 
and 
and 
and 
and 


D 

••sr  COS 


For  point  m',    —xe  =  —x  -f-^-  sin  0        and 


—x  —-^  sin 
6 


and 


For  point  B,      -xe  =  -x  +—  sin  (j>       and 


— yi =  —  y  ~i~  ~TT  cos  <^> 
D 

7/e  =       y  —  —  COS  0 


—  COS  (f> 

D 


D         , 
=     y  — —  cos  (p 

;=       2/  +  «"  COS^> 

D 


—  x—  ~-si 


and        —iji  =  —y  +—  cos  96 


.  (76D) 


322 


KINETIC  THEORY  OF  ENGINEERING  STRUCTURES          CHAP.  XV 


The  general  equations  for  the  moment  influence  lines  of  the  two  kernel  points  may 
then  be  expressed  in  terms  of  the  ordinates  rj  and  coordinates  (x,  y)  of  the  axial  point. 
Thus 

D    ,      A  J  D 


r        / 

-Tioe       |^a       ^ 


COS 


01 


,    •     •     (76E) 


where  the  coordinates  with  their  proper  signs  are  to  be  taken  from  Eqs.  (76D). 


ART.   77.     MAXIMUM   STRESSES   AND    CRITICAL   SECTIONS 

The  maximum  stress  for  any  given  section  is  determined  from  the  moment  influence 
line  for  that  section  by  placing  the  maximum  train  of  live  loads  over  the  positive  portion 
of  the  influence  line.  Similarly  the  minimum  stress  in  found  by  loading  the  negative 
portion  of  the  moment  influence  line. 

Any  point  for  which  the  TJ  ordinate  of  a  moment  influence  line-  is  zero,  must  be  a 
load  divide,  and  there  is  no  satisfactory  way  of  locating  the  load  divides  other  than  by 
drawing  the  influence  lines.  This  will  be  illustrated  in  connection  with  the  problem 
which  follows. 

By  the  method  of  influence  lines  we  are  thus  enabled  to  find  the  maximum  and 
minimum  stresses  for  any  particular  section. 

However,  this  does  not  afford  sufficient  knowledge  regarding  the  safety  of  a  structure 
unless  those  sections  are  examined  which  receive  the  greatest  maximum  stresses  for  the 
most  unfavorable  positions,  of  the  live  load  considering  the  structure  as  a  whole.  These 
sections  of  greatest  maxima  are  called  critical  sections,  and  for  any  given  structure  all  of 


ART.  77  FIXED  MASONRY  ARCHES  323 

the  critical  sections  must  be  investigated  while  all  other  sections  are  of  minor  importance 
and  usually  receive  no  consideration. 

Unfortunately  the  theory  of  elasticity  does  not  afford  a  direct  solution  for  finding 
the  critical  sections.  The  critical  sections  might  be  located  after  a  number  of  sections 
all  along  the  arch  ring  had  been  examined,  and  the  points  of  greatest  maxima  would 
correspond  to  the  critical  sections.  However,  this  would  involve  much  labor  and  from 
the  nature  of  the  case  is  not  warranted,  since  the  stresses  in  an  arch  ring  always  change 
gradually  and  not  abruptly.  An  approximate  knowledge  of  the  location  of  the  critical 
sections  is,  therefore,  all  that  is  required  and  this  may  be  had  from  a  discussion  of  the 
moment  equation. 

Thus  the  general  equation  for  moments  about  any  point  m  by  Eq.  (74s)  is 

Mm  =Mom  -[Xa  +xmXb  +ym  cos  pXc] 

wherein  the  parenthetic  quantity  may  be  positive  or  negative,  depending  on  the  location 
of  the  point  m. 

Since  the  stresses  are  direct  functions  of  Me  and  M{  for  a  given  section,  then  these 
moments  and  stresses  attain  maximum  values  simultaneously: 

1.  For  Xa  negative  and  Xj,  and  Xc  both  positive,  when  ym=0   or   when    xm  =0. 
This  would  locate  three  critical  sections,  one  at  the  crown  for  xm  =0,  and  two  on  the  x 
axis  for  ym=Q. 

2.  For  Xa,  Xb,  and  Xc  all  positive,   Mm  may  become  maximum  when  ym   has   its; 
greatest  negative  value  or  when  both  xm  and  ym  ha-ve  their  greatest  negative  values. 
This  locates  two  other  critical  points,  one  in  each  abutment. 

Hence,  there  are  five  critical  sections  in  every  arch  which  must  be  carefully  examined 
for  maximum  and  minimum  stresses,  and  the  positions  of  the  moving  loads  for  these 
limiting  stresses  are  given  from  the  influences  lines  drawn  for  the  five  critical  sections. 
In  the  case  of  symmetric  arches,  only  three  critical  sections  require  investigation. 

The  two  critical  sections  for  ?/=0  fall  very  close  to  the  quarter  points  of  the  arch, 
a  fact  which  is  important  in  making  preliminary  designs  by  the  method  of  resultant 
polygons.  The  load  divide  may  then  be  approximately  located  in  the  manner  given 
for  three-hinged  arches  by  treating  the  crown  as  a  hinged  point. 

The  maximum  stresses  are  not  appreciably  altered  by  taking  the  quarter  point 
critical  sections  a  little  to  one  side  or  the  other  of  the  real  theoretical  point,  while  the 
stresses  are  greatly  affected  by  the  shape  of  the  axial  line  of  the  arch  ring. 

Thus  for  maximum  economy,  the  axial  line  should  be  a  line  of  zero  moments  for  the 
case  of  average  loading.  That  is  Mm  should  be  zero  for  every  point  of  the  axis  of  the  ring 
when  the  arch  is  carrying  a  dead  load  and  half  the  maximum  live  load  over  the  whole  svan. 


324 


KINETIC  THEORY  OF  ENGINEERING  STRUCTURES 


CHAP.  XV 


ART.  78.     RESULTANT   POLYGONS 

The  resultant  polygon  for  any  case  of  applied  loads  P  may  be  located  by  finding 
the  corresponding  values  of  Xa,  X^,  and  Xc  from  the  three  influence  lines  for  these 
redundants,  having  previously  located  the  (x,  y)  axes. 

The  redundants  Xc  and  Xb,  Fig.  78A,  were  taken  as  the  components  of  H'  ,  respectively 
coincident  with  the  x  and  y  axes.  The  angle  /?,  which  the  x  axis  makes  with  the  hori- 
zontal, is  given  by  Eq.  (73n)  and  the  y  axis  was  taken  vertically.  Hence  the  horizontal 
component  H  ,  of  Xc,  is 

H=Xccos{3      and,    from  Fig.  72A, 


tan  a= 


H'  = 


H 


H 

Xc  cos  /? 
cos  a 


cos  a 

Xa       _Xa 
H'  cos  a  ~ H 


H 

Co  =^!  ^  (tan  a  -tan  /?) 
ti 


Xa=Xcz0  cos  fi=Xbz0- 
c 


(78A) 


These  dimensions  fix  the  closing  line  a\b\  on  the  two  end  verticals  of  the  abutments, 
also  the  haunch  thrust  H'  ,  all  in  terms  of  Xa,  Xb,  and  Xc. 

The  reactions  R\  and  #2  may  be  found  from  the  vertical  reactions  A0  and  B0  and 
the  thrust  H'.  The  vertical  reactions  are  those  due  to  loads  P  acting  on  a  simple  beam 

0  =  ^(l—e)  P/l  and  B0=^Pe/l.  The 


and   may  be  computed  from  Eqs.  (72A)  as 
reactions  R\  and  R%  must  then  intersect  on  the  line  of  the  resultant  R  of  all  the  applied 
loads  P. 

A  force  polygon,  Fig.  78A,  is  drawn  by  laying  off  all  the  loads  P  in  proper  succession, 
dividing  this  load  line  into  the  parts  A0  and  B0  and  at  the  dividing  point  drawing  a  line 
parallel  to  aib\  of  length  equal  to  H'.  This  determines  the  pole  of  the  force  polygon 
from  which  the  reactions  RI  and  R2  and  the  resultant  polygon  through  «i  and  bi  are 
easily  drawn.  See  also  the  force  polygon  in  Fig.  52A,  showing  how  the  pole  0  is  located 
when  H',  A0,  and  B0  are  given. 

The  resultant  polygon  thus  found  must  intersect  the  arch  center  line  at  least  three 
times  for  unsymmetric  loading  on  an  unsymmetric  arch,  and  four  times  for  a  symmetric 
arch,  provided  the  arch  center  line  was  so  chosen  as  to  be  coincident  with  the  resultant 
polygon  drawn  for  the  case  of  average  loading.  This  average  loading  consists  of  the 
total  dead  load,  plus  half  the  live  load  uniformly  distributed  over  the  entire  span.  It 


ART.  78 


FIXED  MASONRY  ARCHES 


325 


will  be  so  understood  whenever  average  loading  is  referred  to  in  connection  with  arch 
designs. 

The  most  favorable  resultant  polygon  is  the  one  which  coincides  with  the  axial  line 
of  the  arch  ring.  Hence  the  most  economic  shape  for  an  arch  is  the  one  for  which  the 
axial  line  is  chosen  to  be  the  resultant  polygon  drawn  for  the  case  of  average  loading 
as  just  defined.  The  live  load,  in  its  critical  positions,  will  then  produce  minimum 
values  for  the  moment  Mm=Nmv  by  making  the  offsets  v,  between  the  axial  line  and 
the  resultant  polygon,  all  minimum.  The  absolute  minimum  for  Mm=Nmv  can  occur 


FIG.  78A. 


omy  for  v=Q  at  all  arch  sections,  and  this  condition  can  be  produced  only  for  one  case 
of  loading,  which  is  taken  as  the  average  loading,  when  maximum  economy  is  to  be 
achieved  in  the  design. 

For  all  cases  of  live  loads  other  than  the  average,  the  moments  Mm--=Nmv  could 
never  become  zero  for  all  points  of  the  axial  line,  though  these  moments  would  be 
minimum  for  an  arch  properly  designed  for  the  average  loading. 


326 


KINETIC  THEORY  OF  ENGINEERING  STRUCTURES 


CHAP.  XV 


ART.  79.     EXAMPLE— 150  FT.   CONCRETE   ARCH 

(a)  Given  data  and  preliminary  design.  The  data.  Required  to  design  a  single 
track  railway  arch  of  150  ft.  clear  span  and  50  ft.  clear  rise  with  profile  of  surface  and 
foundations  as  shown  in  Fig.  79fi.  The  loading  to  be  Cooper's  EGO,  allowing  full 
impact  according  to  formula  (1),  Eqs.  (64c),  300  -r-[l  +300],  where  I  is  the  portion  of 
span  covered  by  loads,  for  any  case  of  loading. 

The  material  shall  be  concrete,  mixed  in  the  proportion  of  one  part  American 
Portland  cement  to  two  parts  sand  to  four  parts  crushed  granite  or  limestone.  Cubes 
of  12  inches  on  each  side,  at  the  age  of  6  months,  shall  sustain  a  compressive  stress 
of  not  less  than  3600  Ibs.  per  sq.  in.  and  briquettes  90  days  old,  mixed  one  part  cement 
to  two  parts  sand,  shall  not  break  below  300  Ibs.  per  sq.  in.  in  tension. 

Accordingly  the  weight  of  one  cu.  ft.  of  concrete  is  assumed  at  140  Ibs.,  and  for 
stresses  between  100  and  600  Ibs.  per  sq.  in.  £"=3,670,000  Ibs.  per  sq.  in.,  and 


13' 


FIG.  79A — Pier  of  Roadway  3'X  13'. 

e  =0.0000054  per  1°  F.  The  allowable  unit  compressive  stress  will  be  500  Ibs.  per  sq. 
in.  or  515  cu.  ft.  of  concrete  per  sq.  ft.  of  surface.  All  pressures  will  be  expressed  in 
cu.  ft.  per  sq.  ft.,  a  unit  which  is  very  convenient  and  nearly  the  same  as  Ibs.  per  sq.  in. 
Thus  if  7--=144  Ibs.,  then  Ibs.  per  sq.  in.  =cu.  ft.  per  sq.  ft. 

Design  of  roadway  It  is  .proposed  to  place  the  track  on  a  floor  built  of  concrete 
and  reinforced  by  rolled  I-beams  of  sufficient  strength  to  carry  the  whole  wheel  loads 
over  spans  10  ft.  in  length.  This  floor  system  is  supported  by  concrete  piers  3  ft.  thick 
and  13  ft.  wide  and  these  in  turn  transmit  the  loads  to  the  arch  ring.  This  floor  is  shown 
in  section  Fig.  79A. 

Each  I-beam  must  carry  two  30,000  Ib.  wheel  loads  spaced  five  feet  apart  and 
producing  a  maximum  bending  moment  of 

M  =30,000X60-30,000X30=900,000  in.  Ibs. 

Allowing  97  per  cent  impact  for  a  10  ft.  span,  this  moment  becomes  1,733,000  in. 
Ibs.,  and  for  an  allowable  unit  stress  of  16,000  Ibs.,  the  required  section  modulus  is 


ART.  79  FIXED  MASONRY  ARCHES  327 

A// 16,000  =  111  in  inches.     One  20-inch  I,  65  Ibs.,  has  a  section  modulus  of  117,  which 
is  ample. 

Weight  of  floor.     2  ft.  of  ballast  at  120  Ibs.  per  cu.  ft. .  .3850  Ibs. 

Concrete '.  .5000   " 

2  I-beams  and  cor.  bars 150   " 

2  rails 200   " 

Total  per  foot,  of  roadway 9200  Ibs. 

Total  per  sq.  ft.  of  arch  ring  q=7lO  Ibs.  =5.07  cu.  ft. 

Live  loads.  E60  loading  for  150  ft.  span,  all  locomotives,  gives  8400  Ibs.  per  ft.  of 
one  track,  and  this,  distributed  over  a  13  ft.  wide  arch,  gives  640  Ibs.  per  sq.  ft.  Adding 
67  per  cent,  impact  for  150  ft.  span,  then  the  total  live  load  p  becomes  1.67x640  =  1070 
Ibs.  per  sq.  ft.  -=7.64  cu.  ft.  concrete  per  sq.  ft. 

Preliminary  design  by  resultant  polygons.  The  methods  outlined  in  Art.  71  are 
applied  here  and  the  design  is  shown  in  Fig.  79s. 

Taking  dimensions  above  given  and  assuming  that  the  stress  for  dead  load  plus  half 
uniform  live  load  will  be  about  0.4  of  the  allowable,  then  Eq.  (71  A)  gives 


. 

ho 

and  Eqs.  (71s)  give  D0=6.00  ft.  which  value  is  accepted  for  the  crown  thickness  of  the 
arch. 

Eq.  (71c)  gives  81.25  ft.  for  the  radius  of  the  intrados  at  the  crown,  and  the  quarter 
point  for  the  parabola  passing  through  the  same  three  points  at  crown  and  springing, 
is  from  Eq.  (7 ID) 

y  J*  =  12.5  feet      and      x  -?  =37.5  feet. 
4  4 

The  point  on  the  parabola  is  plotted  at  s,  Fig.  79s,  and  the  point  s'  is  on  the  circle 
drawn  with  radius  81.25  ft.  The  point  s",  midway  between  s  and  s',  is  chosen  and  a 
three-center  curve  is  drawn  through  s"  and  the  given  crown  and  springing  points. 

The  thickness  of  the  arch  at  the  point  s"  is  found  to  be  6.9  ft.,  using  Eq.  (7lE). 

The  remainder  of  the  ring  is  drawn  according  to  good  judgment  and  the  thickness 
at  the  springing  was  taken  at  10  ft. 

The  arch  ring  is  now  divided  up  into  an  even  number  of  sections  as  described  in 
Art.  73,  so  that  the  same  subdivisions  may  be  used  throughout  the  final  analysis.  The 
dead  loads  are  then  computed  and  tabulated  together  with  the  live  loads  as  required 
for  the  construction  of  the  resultant  polygons.  All  this  is  given  in  Table  79*. 


328 


KINETIC  THEORY  OF  ENGINEERING  STRUCTURES          CHAP.  XV 


m 

Oi 

t> 

d 

h-l 
fe 


ART.  79 


FIXED  MASONRY  ARCHES 


329 


TABLE  79A 
DEAD   LOADS   AND   EQUIVALENT  UNIFORM   LIVE    LOADS 


Ordinates 
V 
from 
.4 

ft. 

Length 
on 
Axis, 

Jw 

ft. 

D 

ft. 

DEAD  LOADS  IN  CUBIC  FEET. 

Live 
Load, 
P 

cu.ft. 

•tf 

cu.ft. 

Q+P 

cu.ft. 

Arch 
Sections, 
sq.ft. 

Piers. 

sq.ft. 

Road- 
way. 

sq.ft. 

Total  Q. 
cu.ft. 

0-  0.0 
1-5.9 
2-12  A 
3-18.9 
4-25.4 
5-31.9 
6-38.4 
7^4.9 
8-51.4 
9-57.9 
10-64.4 
11-71.9 
12  79  4 

10.00 
9.30 
8.60 
8.00 
7.60 
7.20 
6.90 
6.64 
6.46 
6.30 
6.20 
6.10 
6.00 

193.44 
145.40 
115.47 
95.82 
87.33 
91.62 

108.00 
64.80 
42.00 
21.00 
14.80 
6.00 

65.91 
65.91 
65.91 
65.91 
65.91 
76.05 
Totals 

<31  =  367.35 
#3  =  276.11 
Q5  =  223.38 
Q7  =  182.73 
#,  =  168.04 
Qu  =  173.67 

99.32 
99.32 
99.32 
99.32 
99.32 
114.60 

^  =  417.01 
P3  =  325.77 
P5  =  273.04 
P7=232.39 
P9  =  217.70 
Pu  =  230.97 

P'23  =  466.67 
P'n  =  375.43 
P'1B  =  322.70 
P'17  =  282.05 
P'is  =  267.36 
P'13  =  288.27 

20.80 

18.10 

16.00 

14.40 

13.84 

15.02 

1391.28 

611.20 

1696.88 

2002.48 

Above  loads  are  all  for  an  arch  ring  one  foot  thick. 


The  resultant  polygon  for  the  symmetric  loading  Q+^P.  given  in  Table  79A  as  loads  PI 
to  PI  i,  may  now  be  drawn  so  as  to  pass  through  the  center  of  the  arch  ring  at  the  joint 
a  and  at  the  crown  n.  This  polygon  is  shown  in  Fig.  79B,  as  a  fine  line  almost  coincident 
with  the  center  line,  and  the  shape  of  the  arch  ring  is  thus  found  to  be  acceptable.  Had 
this  coincidence  not  been  so  close  then  the  shape  of  the  arch  would  require  modification. 

For  the  above  case  of  symmetric  shape  and  loading,  the  resultant  polygon  for  the 
half  span  only  is  required,  which  is  best  constructed  by  computing  the  horizontal  thrust 
instead  of  finding  it  graphically.  This  is  done  by  computing  the  moments  of  the  forces 
PI  to  PII  about  n  and  dividing  the  sum  of  these  moments  by  #  =  2P  to  obtain  the  lever 
arm  x  of  R.  The  vertical  component  of  the  reaction  at  A  is  equal  to  R  and  hence  H  is 
found  by  taking  moments  about  n  of  all  the  external  forces  to  the  left  of  n.  This  gives 


H=Q 


h 


1696.88(79.4-45.83) 
50.4 


=  1130.2  cu.ft. 


The  force  polygon  is  thus  easily  constructed  and  the  resultant  polygon  is  then  drawn 
through  the  two  section  centers  0  and  the  crown. 

The  resultant  polygon  for  maximum  one-sided  loading,  is  now  constructed.  To  do  this 
the  load  divide  for  the  quarter  point  m  is  found  as  indicated  in  Fig.  79s  by  drawing  lines 
am  and  bn  which  intersect  in  the  required  point  i. 

The  full  load  p  =7.64  cu.ft.  per  foot  is  allowed  to  cover  the  span  from  the  right  up  to 
the  point  i  and  the  left  portion  is  acted  on  by  dead  loads  only.  The  loads  Q\  to  QQ  and 
P'n  to  P'zz,  from  Table  79A,  are  now  united  into  a  force  polygon  with  assumed  pole 


330 


KINETIC  THEORY  OF  ENGINEERING  STRUCTURES 


CHAP.  XV 


distance  E'\  and  an  equilibrium  polygon  a'cb'  is  drawn  for  the  purpose  of  finding  the 
reactions  R\  and  R'-z  and  the  correct  pole  03. 

It  is  then  required  to  pass  a  resultant  polygon  through  the  three  points  a,  n  and 
6,  where  a,  and  b  are  middle  third  points  and  n  is  D/8  above  the  axial  line.  This  is  done 
as  follows:  project  the  points  a,  n  and  b  down  vertically  on  the  equilibrium  polygon, 
giving  the  points  a',  c  and  b'  respectively.  Draw  a'c  and  b'c.  Then  in  the  force  polygon 
make  0>2c'  \\  a'c  and  O^cf'  \\cb'  thus  establishing  the  points  c'  and  c".  Now  draw  the  lines 
c'03  ||  an,  and  c"0s  \\bn  giving  the  intersection  03  as  the  required  pole  and  the  line  U^T  [| ~ah 
will  be  the  true  pole  distance.  An  equilibrium  polygon  drawn  through  the  point  a,  using 
the  force  polygon  OsMN,  will  then  pass  through  the  other  two  points  n  and  b.  The  loads  are 
divided  at  T  into  the  reactions  .4  0=  If  rand  B0=NT&s  per  Eqs.  (72A)  &ndOsM=R'i  and 
03N«*R>3. 

By  the  construction  given  in  Fig.  79s,  the  correct  pole  H'  =1220  cu.ft.,  and  the 
required  resultant  polygon  anb  are  found.  The  polygon  intersects  the  arch  center  line  in 
three  points  and  remains  just  inside  the  middle  third  of  the  arch  ring  at  all  critical  sections. 

It  is  seen  by  inspection  that  a  more  favorable  resultant  polygon  could  not  be  drawn, 
hence  the  solution  is  considered  completed  provided  the  stresses  resulting  from  this  load- 
ing do  not  exceed  the  allowable  limits. 

The  stresses  are  now  found  from  Eq.  (69A)  which,  for  a  rectangular  section  of  depth 
D  and  width  unity,  becomes 


B 


D 


(79B) 


where  v  is  the  lever  arm  of  the  normal  N  about  the  axial  point  of  the  section. 

TABLE  79B 
STRESSES  FOUND   FROM   PRELIMINARY  DESIGN 


Section. 

V 
ft. 

D 

ft. 

UNSYMMETKIC  LOAD  Q  +  P. 

LOADS  Q  +  P  OVER  WHOLE  SPAN. 

N 

cu.ft. 

V 

ft. 

fe 
cu.ft. 

ft 

cu.ft. 

N 
cu.ft. 

V 

ft. 

fe 
cu.ft. 

ft 

cu..t. 

a 

0.0 

10.00 

1980 

1.67 

396.0 

0.0 

2372 

0 

237.2 

237.2 

m 

41.3 

6.77 

1390 

1.13 

0.0 

410.7 

n 

79.4 

6.00 

1220 

0.75 

406.6 

0.0 

1266 

0 

422.  0 

422.0 

- 

94.4 

6.20 

1234 

1.03 

398.0 

0.0 

b 

158.8 

10.00 

2290 

1.67 

0.0 

458.  0 

2372 

0 

237.2 

237.2 

It  is  thus  seen  that  the  stresses  are  all  well  on  the  safe  side  and  the  preliminary 
design  is,  therefore,  acceptable.  Had  it  been  impossible  to  construct  a  resultant  polygon 
for  the  unsymmetric  loading,  which  would  remain  within  the  middle  third  of  the  arch 
ring,  then  the  thickness  of  the  ring  would  have  to  be  altered  accordingly.  Also,  if  the 
unit  stresses  had  exceeded  the  allowable  limits  the  thickness  of  the  arch  would  have  to 
be  increased. 


ART.  79  FIXED  MASONRY  ARCHES  331 

The  structure  thus  designed  and  found  adequate  to  carry  the  required  loads  safely 
is  now  subjected  to  a  rigid  analysis  by  applying  the  foregoing  formulae,  derived  from  the 
theory  of  elasticity. 

(b)  Analysis  of  the  preliminary  design  by  the  theory  of  elasticity  treating  the 
structure  as  symmetric.  The  computations  are  all  given  in  tabular  form,  thus 
illustrating  more  clearly  the  method  of  conducting  such  analyses  and  furnishing  a  record 
of  the  numerical  work  which  is  well  suited  for  reference  and  checking. 

The  computations  for  the  elastic  loads  W,  and  the  location  of  the  (x,  y)  axes  are  carried 
out  in  Table  79c,  following  closely  the  method  described  in  Art.  73. 

The  original  subdivisions  of  the  arch  ring  are  still  retained  and  each  space  is  again 
divided  into  two,  making  24  sections  in  the  whole  ring  between  a  and  b. 

The  programme  carried  out  in  the  table  is  as  follows:  Compute  the  loads  Wa 
from  Eq.  (73A)  then  determine  z0'  and  transform  the  original  (v,  z)  axes  to  the  (x,  y)  axes 
according  to  Eqs.  (73j).  Finally  compute  the  Wb  and  Wc  loads  by  Eqs.  (73A),  and  the 
pole  distances  Ha,  Hb  and  Hc  from  Eqs.  (74A).  This  then  furnishes  all  the  data  for  con- 
structing the  Xa,  Xb  and  Xc  influence  lines. 

The  y  axis  is  known  to  be  the  axis  of  symmetry  and  must,  therefore,  pass  through 
the  crown  joint  n  or  12.  The  location  of  the  x  axis  is  found  by  computing  the  ordinate 
2,,'  of  the  origin  0  from  some  assumed  v  axis  which  was  taken  26.4  ft.  below  the  springing 
points  0  and  24,  Fig.  79u. 

In  Table  79c,  the  dimensions  D,  z  and  sec  <j>  are  tabulated  from  the  drawing  and  from 
these  the  values  I/I,  wa  and  zwa  are  derived.  The  abscissa;  x  and  increments  Ax  are  also 
taken  from  the  drawing,  and  the  other  values  result  from  performing  the  operations 
indicated  in  the  headings.  It  should  be  noted  that  the  constant  modulus  E  was  not 
considered  in  computing  the  elastic  loads  W  so  that  all  these  loads  are  E  times  actual. 

The  secants  are  computed  from 

R  115.8    , 

sec  0  =  — 7^-^  =  — ^r^~  f or  Pomts  between  a  and  ra, 

2+OO.O        2  +00.;) 

sec  &  7=  — '^-^  for  points  between  m  and  n. 
2+0.9 

The  sums  '£wa  =  Wa,  2zwa=zTF0,  etc.,  are  computed  by  Simpson's  rule  for  distance 
increments  Ax,  according  to  Eqs.  (73 E). 

The  ordinate  z0' ,  of  the  center  of  gravity  or  origin  of  coordinates,  is  then  found  to 
be  22TFa/2PFa=64.72  ft.  Also,  the  sum  2zzTFa=0  when  extended  over  the  whole 
arch,  hence  tan  /3=0,  making  y=z—z0'. 

The  functions  xwa,  x2wa,  ywa  and  y2wa  are  now  figured  and  from  these  the  sums 
2.ru?  =Wb,  H>x2wn=xWb,  Hywn  =  Wc  and  ^y2wa=yWc  are  again  derived  by  applying 
Simpsons's  rule  for  the  distance  increments  Ax.  It  should  be  noted  that  for  the  whole 
arch  ring  2  JF&  =0  and  2JF  =0,  serving  as  a  check  on  the  computations.  If  these  sums  d6 
not  become  zero,  then  errors  have  occurred  which  must  be  rectified.  Small  discrepancies 
will  always  be  found  owing  to  insufficient  accuracy  in  the  data  and  not  carrying  sufficient 
decimals.  These  may  be  adjusted  by  proportionate  distribution.  However,  this  must 


332 


KINETIC  THEORY  OF  ENGINEERING  STRUCTURES 


CHAP.  XV 


GO 

w 
o 


c 

PH 
C 


o 

b 


05       *< 

*~    w  o 


s  tf 

3  H 

^  H 
N| 

^  02 

w 
i 

EH 

fe 
O 

K 

o 

I— I 

I 


•? 

o 

s 

H 

10 

CH 

§ 

O 


,s> 

"si  >i 


•6- 


C5COOOr(H>-H 


IOOC5OOO1—  I    i-l    i—  l 


+ 


ooooooooooooo 

I    I    I    I    I    I  + 


1  + 


?O(MOC5t-HCDGOTtl'Ttl'—  IT-HT^T-  1 


i—  I    O5    iO    CC    CO    LO    Cl    i—  i    CO 


(N(N(N(MIM(MGOGOOOOOOOXC» 


U5    t>-    O    i—  i 


Oi-Hi-HlMIMfNINIMCOCOCOeOIN 
OOOOOOOOOOOOO 


co  co  co  co  co  co 


pppooooooopop 
ooooooooooooo 


-©- 


ooppppppppppp 
odo'do'o'ooooo'oo 


I  •* 

N       O 

II  -S 

55  -I 

M 


was  omil 


o 
H 


ART.  79 


FIXED  MASONRY  ARCHES 


333 


be  confined  to  small  errors  not  exceeding  ^  per  cent.     Mistakes  must  be  corrected  and  not 
distributed. 


Lisa's 


FIG.  79c. — X  Influence  Lines  for  "Symmetric  Arch. 

The  influence  lines  for  Xa,  X6,  and  Xc  were  drawn  in  Fig.  79c,  using  the  results  from 
Table  79c,  and  neglecting  the  effect  Scos  (j>Jx/EF  due  to  the  axial  thrust  N  on  the  pole 
distance  Hc  of  the  Xc  influence  line.  If  this  effect  is  to  be  considered  then  the  value 
£*0  cos  (f>4x/EF  would  be  determined  in  the  manner  later  indicated  in  Table  79M,  omitting 


334 


KINETIC  THEORY  OF  ENGINEERING  STRUCTURES 


CHAP.  XV 


the  constant  E,  since  it  was  also  omitted  in  computing  the  values  wa.  This  is  permissible 
because  E  being  involved  as  a  reciprocal  factor  in  all  elastic  loads,  as  per  Eqs.  (72o), 
may  be  considered  as  canceled  in  Eqs.  (72ii).  It  must  be  remembered,  however,  that  the 
elastic  loads  are  then  actually  E  times  too  large,  a  point  which  must  be  considered  in 
computing  temperature  effects  by  Eqs.  (7oB). 

The  X  influence  lines  are  the  equilibrium  polygons  for  the  W  loads  when  certain 
pole  distances  are  employed  in  the  respective  force  polygons.  It  is  generally  convenient 
to  draw  these  diagrams  so  that  the  influence  line  ordinates  may  appear  a  certain  number 
of  times  actual  when  scaled  to  the  scale  of  lengths  of  the  drawing. 

The  Xa  influence  line  is  the  equilibrium  polygon  for  the  Wa  loads  when  the 
pole  distance  is  made  equal  to  Ha  =  HWa  =6.564,  as  given  by  Table  79c.  In  the 
present  case  the  pole  was  made  ^£TFa  so  that  the  ordinates  >ja  of  the  Xa  influence 


FOR  INTRADOS  KERNEL  POINTS. 
37/0         m  n 


B 


FIG.  79o. — Mo  Influence  Lines  for  Symmetric  Arch. 

line  become  twice  actual  to  the  scale  of  lengths.  The  scale  of  the  force  polygon  is 
chosen  so  as  to  give  a  reasonably  large  figure  and  to  permit  of  accurate  construction 
of  the  equilibrium  polygon,  otherwise  the  scale  of  loads  need  not  bear  any  specified 
relation  to  the  scale  of  ordinates.  As  a  check  on  the  drawing,  the  outer  rays  of  this 
influence  line  must  intersect  on  the  y  axis,  in  fact  this  is  the  graphic  method  of  locating 
this  axis. 

The  Xb  influence  line  is  the  equilibrium  polygon  for  the  Wb  loads  when  the  pole  dis- 
tance is  made  equal  to  Hb  =  2xWb  =2X5482.9  =  10965.8,  from  Table  79c.  As  this  pole 
is  very  large  in  comparison  with  the  2^  =  ±107.4,  it  is  advisable  to  take  Hb=Q.QlHxWb 
=  109.66,  making  the  ordinates  rjb  one  hundred  times  actual  to  the  scale  of  lengths.  The 
scale  of  the  Wb  loads  was  again  chosen  as  a  matter  of  convenience.  The  closing  line 
AB  divides  the  influence  line  into  a  positive  and  a  negative  area. 

The  Xc  influence  line  is  the  equilibrium  polygon  for  the  Wc  loads  when  the  pole 
is  made  equal  to  #c=cos  8^yWc  =2X560.9  =  1121.8.  Here  also,  the  pole  is  too  large 


ART.  79 


FIXED  MASONRY  ARCHES 


335 


to  furnish  a  well  shaped  force  polygon  and  hence  the  pole  was  made  #C=0.02£?/TFC, 
giving  influence  line  ordinates  >jc  fifty  times  actual  to  the  scale  of  lengths.  The  scale 
of  Wc  loads  is  also  chosen  as  a  matter  of  convenience. 

The  M0  influence  lines  are  ordinary  moment  influence  lines  drawn  for  the  simple  span 
AB  and  the  several  kernel  points  of.  the  critical  sections.  Fig.  79o.  The  ordinates  may 
be  made  actual  to  the  scale  of  lengths  or,  as  in  the  case  of  the  Mot  line,  the  ordinates  were 
made  five  times  actual. 

The  M;  and  Me  influence  lines  for  the  kernel  points  of  the  critical  sections  are  drawn 
by  plotting  the  coordinates  computed  according  to  Eqs.  (76E).  See  Fig.  79E  and 
Tables  79E  and  79F. 

The  coordinates  of  the  kernel  points  i  and  e  of  the  three  critical  sections  at  t,  m  and  n 
of  the  arch  ring,  are  computed  from  the  coordinates  (x,  y)  of  the  axial  points,  using  Eqs. 
(76o)  as  given  in  Table  79D. 

TABLE  79D 
DATA   RELATING   TO   THE    CRITICAL   SECTIONS 


AXIAL  POIXT. 

KERNEL,  POINTS. 

D   . 

n 

Point. 

D 

* 

—  sin  c£ 

—    COS  <£ 

X 

V 

xi 

Vi 

xe 

Ve 

ft. 

ft. 

ft. 

ft 

ft. 

ft, 

ft, 

ft. 

ft. 

t 

9.75 

56°  30' 

78.05 

-36.19 

1.36 

0.89 

79.4 

-35.3 

76.7 

-37.1 

m 

6.90 

30°  30' 

41.8 

0.0 

0.62 

0.99 

42.4 

+   0.99 

41.2 

-   0.99 

n 

6.00 

0°  00' 

0.0 

+  12.08 

0.00 

1.00 

0.0 

+  13.08 

0.0 

+  11.08 

From  the  coordinates  of  the  kernel  points  Table  79o,  and  the  ordinates  in  Tables 
79E  and  79r,  taken  from  the  Xa,  Xb,  Xc  and  M0  influence  lines,  t)e  and  TH  are  computed 
for  each  section  of  the  span  and  for  the  critical  sections  t,  m  and  n. 

Thus,  for  the  section  at  m,  Eqs.  (76B)  may  be  written 


(79c) 


The  computations  are  all  indicated  in  the  headings  of  the  tables  and  no  further 
comment  is  necessary  here. 

With  the  use  of  Mf  and  Me  influence  lines  for  the  three  critical  sections  of  a  symmetric 
arch,  the  stresses  in  these  sections  may  be  found  for  any  possible  case  of  loading. 

The  resultant  polygon  for  any  particular  case  of  simultaneous  loading,  is  located 
when  the  redundants  A"a,  Xb  and  Xc  are  known.  Eqs.  (78A)  then  give  the  necessary 
dimensions  from  which  the  resultant  polygon  may  be  drawn. 

Table  79o  gives  the  computations  for  the  resultant  polygon  for  the  case  of  average 
loading  Q  +%P,  over  the  entire  span.  This  is  the  case  for  which  the  resultant  polygon 
should  coincide  with  the  axis  and  the  results  are  comparable  with  those  on  Fig.  79s. 


336 


KINETIC  THEORY  OF  ENGINEERING  STRUCTURES 


CHAP.  XV 


The  lettered  dimensions  in  the  table  are  all  shown  in  Fig.  7SA,  but  the  resultant  polygon 
is  not  drawn  for  this  case. 

The  influence  line  ordinates  ya,  TJ&  and  TJC  are  taken  from  Fig.  79c  and  the  sums  of  the 
products  of  the  loads  and  ordinates  give  the  respective  redundants  X  from  which  z0,c 
and  c0}  also  A0  and  B0  are  readily  found  by  Eqs.  (78A). 

TABLE  79s 
MOMENT  INFLUENCE   LINES   FOR  EXTRADOS   KERNEL   POINTS.— SYMMETRIC   ARCH 


Pt. 

r]a 
ft. 

r/b 
ft. 

f)c 

ft. 

SECTION  t 
J2«=>?o  —  [i}a+76.7i}6  —  37.l7jc] 

SECTION  m. 

T)m  =  ijo  -  [>?a+  4  1  .2-fjb  -  0.99r;c] 

SECTION  n. 

T)n=T)o  —  [ija+ll.l'jc] 

90 

76.7  T)b 

—  37.lTjc 

fit 

?« 

41.27JC 

0.99i?c 

Tim 

Vo 

H.lTJc 

9» 

0 

0.0 

0.0 

0.0 

0.0 

0.0 

0.0 

0.0 

0.0 

0.0 

0.0 

1 

2.9 

0.031 

0.014 

2.61 

2.38 

-   0.52 

-2.15 

4.5 

1.28 

-0.01 

0.33 

2.9 

0.14 

-0.14 

2 

6.0 

0.062 

0.056 

9.4 

2.55 

-0.05 

0.90 

6.3 

0.62 

-0.32 

3 

8.8 

0.086 

0.120 

2.39 

6.59 

-   4.45 

-8.55 

14.2 

3.54 

-0.12 

1.98 

9.5 

1.33 

-0.63 

4 

11.4 

0.102 

0.200 

19.1 

4.20 

-0.20 

3.70 

12.8 

2.22 

-0.82 

5 

13.9 

0.110 

0.296 

2  17 

8.44 

-10.98 

-9.19 

24.0 

4.53 

-0.29 

5.86 

16.0 

3.29 

-1.19 

6 

16.0 

0.111 

0.396 

29.0 

4.57 

-0.39 

8.82 

19.3 

4.40 

-1.10 

7 

17.8 

0.105 

0.496 

1.95 

8.05 

-18.40 

-5.50 

27.5 

4.33 

-0.49 

5.86 

22.6 

5.51 

-0.71 

8 

19.4 

0.095 

0.588 

25.9 

3.91 

-0.58 

3.17 

25.9 

6.53 

-0.03 

9 

20.7 

0.079 

0.672 

1.73 

6.06 

-24.93 

-0.10 

24.2 

3.25 

-0.67 

0.92 

29.1 

7.46 

0.94 

10 

21.7 

0.058 

0.734 

22.7 

2.39 

-0.73 

-0.66 

32.3 

8.15 

2.45 

11 

22.4 

0.030 

0.782 

1.48 

2.30 

-29.01 

+  5.69 

21.0 

1.24 

-0.77 

-1.87 

36.0 

8.68 

4.92 

12 

22.6 

0.000 

0.800 

19.2 

0.00 

-0.79 

-2.70 

40.0 

8.88 

8.52 

13 

22.4 

-0.030 

0  782 

1  23 

-2.30 

-29.01 

10.14 

17.4 

-1.24 

-0.77 

-2.99 

36.0 

8.68 

4.92 

14 

21.7 

-0.058 

0  734 

15.6 

-2.39 

-0.73 

-2.98 

32.3 

8.15 

2.45 

15 

20.7 

-0.079 

0.672 

0.98 

-6.06 

-24.93 

11.27 

14.0 

-3.25 

-0.67 

-2.78 

29.1 

7.46 

0.94 

16 

19.4 

-0.095 

0.588 

12.4 

-3.91 

-0.58 

-2.51 

25.9 

6.53 

-0.03 

17 

17.8 

-0.105 

0.496 

0.77 

-8.05 

-18.40 

9.42 

10.8 

-4.33 

-0.49 

-2.18 

22.6 

5.51 

-0.71 

18 

16.0 

-0.111 

0.396 

9.3 

-4.57 

-0.39 

-1.74 

19.3 

4.40 

-1.10 

19 

13.9 

-0.110 

0.296 

0.55 

-8.44 

-10.98 

6.07 

7.7 

-4.53 

-0.29 

-1.38 

16.0 

3.29 

-1.19 

20 

11.4 

-0.102 

0.200 

6.1 

-4.20 

-0.20 

-0.90 

12.8 

2.22 

-0.82 

21 

8.8 

-0.086 

0.120 

0.32 

-6.59 

-  4.45 

2.56 

4.6 

-3.54 

-0.12 

-0.54 

9.5 

1.33 

-0.63 

22 

6.0 

-0.062 

0.056 

3.1 

-2.55 

-0.05 

-0.30 

6.3 

0.62 

-0.32 

23 

2.9 

-0.031 

0.014 

0.10 

-2.38 

-   0.52 

0.10 

1.5 

-1.28 

-0.01 

-0.11 

2.9 

0.14 

-0.14 

24 

0.0 

0.0 

0.0 

0.0 

0.0 

0.0 

0.0 

0.0 

0.0 

0.0 

All  ordinates  ij  in  above  table  are  expressed  in  feet. 


Table  79n  presents  the  case  of  unsymmetric  loading  shown  in  Fig.  79s.  The  results 
ere  plotted  in  Fig.  79r,  and  the  resultant  polygon  is  finally  drawn  in  accordance  with  the 
method  given  in  Art.  78. 

The  arch  being  symmetric  /3=0  and  Zi  =£2  hence  with  the  values  of  Xa,  Xb  and  Xc 
from  Table  79n,  the  resulting  data  in  that  table  were  found. 

In  Fig.  79r  the  arch  ring  and  the  (x,  y]  axes  are  given  to  locate  the  points  a  and  b 
on  the  verticals  through  the  end  supports  and  then  to  draw  the  resultant  polygon. 


ART.  79 


FIXED  MASONRY  ARCHES 


337 


/\ 


'    I      2 


\ 


\ 


\ 


\ 


AT  SECTION  t. 
— 1-158'.8  


8    \>    10    II       U     l»      14    IS    16    17     18    >?    30   21    22 


B 


LENGTHS. 

Ijllll.M.I  I  I          __l___L_— 1| 

10         0         IO        20      30       40       SOFT. 
OROI  NATES. 


I// 


itifrados  kernel  points. 
extrados  kernel  points. 


FIG.  79E. — M  Influence  Lines  for  the  Kernel  Points.     Symmetric  Arch. 


338 


KINETIC  THEORY  OF  ENGINEERING  STRUCTURES.          CHAP.  XV 


TABLE  79F 
MOMENT   INFLUENCE    LINES   FOR    INTRADOS   KERNEL    POINTS.— SYMMETRIC  ARCH 


Pt. 

Tja 
ft. 

T,b 
ft, 

1)c 

ft. 

SECTION  t. 
rj  =  ^,0  —  [rja  +  79.4,6  —  35.  3,c] 

SECTION  m. 
,m  =  ,o-  [,3+42.4,6+0.99,0] 

1 
SECTION  n 

T)n=T)o  —  [,<z  +  13.  Irjc] 

Vo 

79.4,6 

—  35.3,6 

yt 

rjo 

42.4,6 

0.99,c 

,m 

9 

13.1,0 

9» 

0 

0.0 

0.0 

0.0 

0.0 

0.0 

0.0 

0.0 

0.0 

0.0 

0.0 

1 

2.9 

0.031 

0.014 

0.0 

2.46 

-   0.49 

-  4.87 

4.5 

1.31 

0.01 

0.28 

2.9 

0.18 

-0.18 

2 

6.0 

0.062 

0.056 

9.4 

2.63 

0.05 

0.72 

6.3 

0.73 

-0.43 

3 

8.8 

0.086 

0.120 

0.0 

6.83 

-   4.24 

-11.39 

14.3 

3.65 

0.12 

1.73 

9.5 

1.57 

-0.87 

4 

11.4 

0.102 

0.200 

19.3 

4.32 

0.20 

3.38 

12.8 

2.62 

-1.22 

5 

13.9 

0.110 

0  296 

0.0 

8.73 

-10.45 

-12.18 

24.2 

4.66 

0.29 

O  .  oC 

16.0 

3.88 

-1.78 

6 

16.0 

0.111 

0.396 

28.1 

4.71 

0.39 

7.00 

19.3 

5.19 

-1.89 

7 

17.8 

0.105 

0.496 

o.c 

8.34 

-17.51 

-  8.63 

26.6 

4.45 

0.49 

3.86 

22.6 

6.50 

-1.70 

8 

19.4 

0.095 

0.588 

25.0 

4.03 

0.58 

0.99 

25.9 

7.70 

-1.20 

9 

20.7 

0.079 

0.672 

0.0 

6.27 

-23.72 

-   3.25 

23.5 

3.35 

0.67 

-1.22 

29.1 

8.80 

-0.21 

10 

21.7 

0.058 

0  734 

22.0 

2.46 

0.73 

-2.89 

32.3 

9.61 

0.99 

11 

22.4 

0.030 

0.782 

0.0 

2.38 

-27.60 

+   2.82 

20.3 

1.27 

0.77 

-4.14 

36.0 

10.24 

3.36 

12 

22.6 

0.000 

0.800 

18.6 

0.00 

0.79 

-4.79 

40.0 

10.48 

6.92 

13 

22.4 

-0.030 

0.782 

0.0 

-2.38 

-27.60 

7.58 

16.8 

-1.27 

0.77 

-5.10 

36.0 

10.24' 

3.36 

14 

21.7 

-0.058 

0.734 

15.1 

-2.46 

0.73 

-4.87 

32.3 

9.61 

0.99 

15 

20.7 

-0.079 

0.672 

0.0 

-6.27 

-23.72 

9.29 

13.6 

-3.35 

0.67 

-4.42 

29.1 

8.80 

-0.21 

16 

19.4 

-0.095 

0.588 

12.0 

-4.03 

0.58 

-3.95 

25.9 

7.70 

-1.20 

17 

17.8 

-0.105 

0.496 

0.0 

-8.34 

-17.51 

8.05 

10.6 

-4.45 

0.49 

-3.24 

22.6 

6.50 

-1.70 

18 

16.0 

-0.111 

0.396 

9.1 

-4.71 

0.39 

-2.58 

19.3 

5.19 

-1.89 

19 

13.9 

-0.110 

0.296 

0.0 

-8.73 

-10.45 

5.28 

7.5 

-4.66 

0.29 

-2.03 

16.0 

3.88 

-1.78 

20 

11.4 

-0.102 

0.200 

6.0 

-4.32 

0.20 

-1.28 

12.8 

2.62 

-1.22 

21 

8.8 

-0.086 

0.120 

0.0 

-6.83 

-   4.24 

2.27 

4.4 

-3.65 

0.12 

-0.87 

9.5 

1.57 

-0.87 

22 

6.0 

-0.062 

0  056 

3.0 

-2.63 

0.05 

-0.42 

6.3 

0.73 

-0.43 

23 

2.9 

-0.031 

0.014 

0.0 

-2.46 

-   0.49 

0.05 

1.4 

-1.31 

0.01 

-0.20 

2.9 

0.18 

-0.18 

24 

0.0 

0.0 

0.0 

0.0 

0.0 

0.0 

0.0 

0.0 

0.0 

0.0 

All  ordinates  in  above  table  are  expressed  in  feet. 

The  construction  is  made  by  laying  off  z0  down  from  the  origin  0  on  the  y  axis  to 
obtain  the  point  s.  Then  lay  off  —  c0  up  from  s  to  determine  s'  on  the  y  axis.  Through 
s'  draw  a  line  parallel  to  the  x  axis  and  where  this  line  intersects  the  vertical  at  A  gives 
the  point  a  on  the  resultant  R'i.  The  line  as  prolonged  to  intersect  the  vertical  at  B 
gives  the  point  6  on  the  resultant  R'2.  Knowing  A0,  B0  and  H' ,  the  resultants  R'\  and 
R*2  can  now  be  drawn,  and  their  intersection  must  fall  on  the  resultant  R,  distant  r  from 
the  2  axis. 

The  force  polygon  is  constructed  by  laying  off  the  load  line  MN  and_then  finding  T, 
which  is  the  point  of  division  between  A0  and  B0.  Through  T  draw  0\T\\ob  and  lay 
off  the  force  H' ,  thus  locating  the  pole  Oi,  and  giving  the  reactions  R'i  and  R'2.  The 
equilibrium  polygon  drawn  through  a  and  with  pole  0\  will  then  pass  through  b  as  a  final 
check.  The  resultant  polygon  for  unsymmetric  loading  intersects  the  axial  line  three 
times,  as  it  should. 


ART.  79 


FIXED  MASONRY  ARCHES 


339 


o 


fe 

£ 

OQ 


w 

o 

OH 

HM 
O" 

CO 
C3       Q 

p    ^ 
o 

H  >-}    ! 

rl 

«          tf 

H     g 


O 

O 


OQ 

3 
3 


fe 
O 

HH 

H 


I 


RESULTING  DATA. 

d 

II 
« 

H 

SB 

+3 

II 
^2. 

W 

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^     » 

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w    II 
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II   ^ 
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u 
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<?       £       *; 

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ftf 

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1^-     OS     Tt<     Tfi     O5     O 
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XI    ^H     00 

05        0 

N" 

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1C     CO    O    O     CO    ^C 
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2  II 

0 

1—4 

ft.|«            0 

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O    O    CO    CO    O    O    ( 
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n  o  O5 

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1        1 

rH      rH     >/? 
+        1       * 

co   o   r^ 

t~-      TjH      Tfl      Ttl      •*      t^      1 

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340 


KINETIC  THEORY  OF  ENGINEERING  STRUCTURES 


CHAP.  XV 


B 

C5 


- 


II 


14 


gO1 


O  ^J 


ti 


-^         co 
CO 

2      8 

co        cs 

00  ^H 


+ 


o 

n.  <H 

oo  •* 

CO    I-H 
(N 


co  o 


a  i 

R 


ft! 


03 


05     -3 


CO"3<N(NOOCOCOOOcOiCTt<  OS 

OOCOCON.OOCOOO3OOOOO  CM 

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00  OS  CO  1C  CD  C^  CO  C^l  »-H  C^ 
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1-Hl-H(N<N(N(Ml-Hl-H 
rl     CO     1C     I>     OS     i— I     CO     1C     f>-     OS     T-l     CO 


ART.  79 


FIXED  MASONRY  ARCHES 


341 


342 


KINETIC  THEORY  OF  ENGINEERING  STRUCTURES 


CHAP.  XV 


The  moments  Me  and  M,  and  the  resulting  stresses  on  any  critical  section,  and  for  any 
case  of  loading,  may  be  found  from  the  moment  influence  lines  Fig.  79E,  drawn  for  the 
kernel  points  of  the  critical  sections. 

The  stresses  for  the  two  cases  of  loading,  Q+^P  and  the  one-sided  load  Q+P,  are 
computed  in  Tables  79i  and  79J,  using  the  influence  line  ordinates  from  Tables  79E  and 
79r,  or  from  the  diagrams  Fig.  79E.  At  the  bottom  of  the  tables  the  values  R,  N,  v,  ig 
and  fi  are  computed  for  each  section,  using  Eqs.  (76c) . 

The  sections  t  and  t',  at  the  abutments,  are  so  located  that  the  intrados  kernel  points 
i  fall  on  the  verticals  through  the  end  axial  points,  thus  avoiding  negative  TJO  values  in 
Eqs.  (79c),  and  other  difficulties  affecting  the  X  influence  lines. 

It  is  seen  from  Table  79i,  that  the  actual  resultant  polygon  for  the-  case  of  average 
loading  really  coincides  very  closely  with  the  arch  center  line  as  designed.  The  offsets 
v  being  +  and  —  show  that  the  polygon  intersects  the  center  line  four  times,  as  it  should. 


MOMENTS   Me  AND   Mi  FOR    THE  CRITICAL    SECTIONS,   LOADS  Q+- 

SYMMETRIC  ARCH 


Loads 

SECTION  I. 

SECTION  m. 

SECTION"  n. 

Pt. 

>S 

1e 

,. 

Me 

Mi 

9« 

ti 

Me 

Mi 

1* 

T) 

Me 

Mi 

cu.ft. 

ft. 

ft. 

ft. 

ft. 

ft. 

ft. 

1 

417.0 

-2.15 

-  4.87 

-  896.6 

-2030.8 

0.33 

0.28 

137.6 

116.8 

-0.14 

'-0.18 

-     58.4 

75.1 

3 

325.8 

-8.55 

-11.39 

-2785.6 

.-3710.9 

1.98 

1.73 

645.1 

563.6 

-0.63 

-0.87 

-   205.3 

-   283.4 

5 

273.0 

-9.19 

-12.18 

-2508.9 

-3325.1 

5.86 

5.35 

1599.8 

1460.6 

-1.19 

-1.78 

-   324.9 

-   485.9 

7 

232.4 

-5.50 

-  £.63 

-1278.2 

-2005.6 

5.86 

3.86 

1366.9 

897.1 

-0.71 

-1.70 

-    165.0 

-   395.1 

9 

217.7 

-0.10 

-   3.25 

21.8 

-  707.5 

0.92 

-1.22 

"200.3 

-   265.6 

4-0.94 

-0.21 

4-   204.6 

-     45.7 

11 

231.0 

4-5.69 

+   2.82 

4-1314.4 

4-  651.4 

-1.87 

-4.14 

-   432.0 

-   956.3 

4.92 

4-3.36 

1136.5 

+   776.2 

13 

231.0 

10.14 

7.58 

2342.3 

1751.0 

-2.99 

-5.10 

-690.7 

-1178.1 

4.92 

4-3.36 

1136.5 

4-   776.2 

15 

217.7 

11.27 

9.29 

2453.5 

2022.4 

-2.78 

-4.42 

-   605.2 

-   962.2 

4-0.94 

^0.21 

+   204.6 

45.7 

17 

232.4 

9.42 

8.05 

2189.2 

1870.8 

-2.18 

-3.24 

-   506.6 

-   753.0 

-0.71 

-1.70 

-    165.0 

-   395.1 

19 

273.0 

6.07 

5.28 

1657  .  1 

1441.4 

-1.38 

-2.03 

-   376.7 

-   554.2 

-1.19 

-1.78 

-   324.9 

-   485.9 

21 

325.8 

2.56 

2.27 

834.0 

739.6 

-0.54 

-0.87 

-    175.9 

-    283.4 

-0.63 

-0.87 

-   205.3 

-   283.4 

23 

417.0 

0.10 

0.05 

41.7 

20.9 

-0.11 

-0.20 

-     45.9 

83.4 

-0.14 

-0.18 

-      58.4 

75.1 

R  = 

3393.5 

D= 

9'.  75 

4-3341.1 

-3282.4 

D= 

6'.  90 

4-1116.7 

-1998.1 

Z>  = 

6'.  00 

4-1175.0 

-1018.0 

N= 

2037.9 

N— 

1354.3 

JV  = 

1096.5 

v= 

4-0'.  014 

v= 

-0'.325 

v  = 

4-0'.  071 

/= 

-   210.9 

-   207.2 

f= 

-    140.8 

-   251.8 

J~ 

-    195.9 

-    169.7 

Eqs.  (76c), 


f  O'M  e 

/e=-^  = 


, 
* 


v  1S  positive  when  measured  from  the  axis 


2N  D*  '         *      W  ' 

toward  the  extrados. 

All  loads  in  above  table  are  expressed  in  cubic  feet  of  masonry  and  the  stresses  are  cubic  feet  per  square  foot.     Cubic 
feet  per  square  foot  X  0.972  =  pounds  per  square  inch. 

The  stresses  in  Table  79j  are  those  due  to  the  case  of  loading  shown  in  Fig.  79r,  and 
hence  the  offsets  v  are  those  which  the  resultant  polygon  in  that  figure  actually  present. 
The  smallness  of  the  scale,  however,  prevents  a  very  close  agreement.  The  resultant 
polygon  for  the  unsymmetric  loading  is  seen  to  intersect  the  center  line  three  times  as 
it  should,  since  two  values  v  are  negative  and  three  positive. 


ART.  79 


FIXED  MASONRY  ARCHES 


343 


TABLE  79j 

MOMENTS  Me  AND  M»  FOR  CRITICAL  SECTIONS.      TOTAL  DEAD  AND  ONE-SIDED  LIVE 

SYMMETRIC  ARCH 


SECTION  I. 

SECTION  m. 

SECTION  n. 

Ft. 

Loads 

Q  and  P', 

i 

rje 

fji 

Me 

Mi 

Tje 

T)i 

Me 

M{ 

fje 

rji 

Me     \      Mi 

cu.ft. 

ft. 

ft. 

ft. 

ft. 

ft. 

ft. 

1 

Qi  =  367.4 

-   2.15 

-   4.87 

-    789.9 

-1789.2 

0.33 

0.28 

121.2 

102.9 

-0.14 

-0.18 

51.4 

-   66.1 

3 

Q3=  276.1 

-   8.55 

-11.39 

-2360.7 

-3144.8 

1.98 

1.73 

546.7 

477.6 

-0.63 

-0.87 

-    173.9 

-  240  .  2 

5 

Ob  =  223.4 

-   9.19 

-12.18 

-2053.0 

-2721.0 

5.86 

5.35 

1309  .  1 

1195.2 

-1.19 

-1.78 

-   265.8 

-397.6 

7 

Qi=  182.7 

-   5.50 

-   8.63 

-1004.9 

-1576.7 

5.86 

3.86 

1070.6 

705.2 

-0.71 

-1.70 

-    129.7 

-310.6 

9 

Qo=  168.0 

-  0.10 

-   3.25 

-      16.8 

-   546.0 

0.92 

-1.22 

154.6 

-   205.0 

+  0.94 

-0.21 

+    157.9 

-   35.3 

11 

Pn'  =  288.3 

+   5.69 

+   2.82 

+  1640.4 

+  813.0 

-1.87 

-4.14 

-539.1 

-1193.6 

4.92 

+  3.36 

1418.4 

+  968.7 

13 

Pis'  =  288.3 

10.14 

7.58 

2923.4 

2185.3 

-2.99 

-5.10 

-862.0 

-1470.3 

4.92 

+  3.36 

1418.4 

+  968.7 

15 

Pis'  =  267.3 

11.27 

9.29 

3012.5 

2483  .  2 

-2.78 

-4.42 

-  743  .  1 

-1181.5 

+  0.94 

-0.21 

+   252.3 

-   56.1 

17 

PIT'  =  282.0 

9.42 

8.05 

2656.4 

2270.1 

-2.18 

—  3.24 

-614.8 

-   913.7 

-0.71 

-1.70 

-   200.2 

-479.4 

19 

P19'  =  322.7 

6.07 

5.28 

1958.8 

1703.8 

-1.38 

-2.03 

-445.3 

-   653.0 

-1.19 

-1.78 

-   384.0 

-574.4 

21 

P2i'=  375.4 

2.56 

2.27 

961.0 

852.2 

-0.54 

-0.87 

-202.7 

-   326.6 

-0.63 

-0.87 

-   236.5 

-326.6 

23 

Pi/  =  466.7 

0.10 

0.05 

46.0 

23.3 

-0.11 

-0.20 

-   51.3 

-     93.3 

-0.14 

-0.18 

-     65.3 

-   84.0 

*- 

3508.3 

D= 

9'.75 

+  6973.9 

+   453.2 

D= 

6'.90 

-256.1 

-3556.1 

D  = 

e'.oo 

+  1740.2 

-632.9 

N= 

2006.4 

N= 

1434.8 

N= 

1186.6 

v= 

+  1'.851 

v  = 

-1'.328 

v  = 

+0'.467 

/= 

-   440.2 

t28.5 

/= 

+     32.3 

-   448.2 

!= 

-   290.0 

-105.5 

Ft. 

Loads 
Q  and  P', 

cu.ft. 

SECTION  m'. 

SECTION  V. 

ft. 

ft. 

Me 

Mi 

ft. 

ft. 

Me 

Mi 

1 

Qi  =  367.4 

-0.11 

-0.20 

•     40.4 

-     73.5 

0.10 

0.05 

36.7 

18.3 

3 

Os  =  276.1 

-0.54 

-0.87 

-    149.1 

-   240.2 

2.56 

2.27 

706.8 

626.7 

5 

#6  =  223.4 

-1.38 

-2.03 

-   308.3 

-   453.5 

6.07 

5.28 

1356.0 

1179.6 

7 

Q7  =  182.7 

-2.18 

-3.24 

-   398.3 

-   591.9 

9.42 

8.05 

1721.0 

1470.7 

9 

Q9=  168.0 

-2.78 

-4.42 

-   467.0 

-    742.6 

11.27 

9.29 

1893.4 

1560.7 

11 

Pii'  =  288.3 

-2.99 

-5.10 

-   862.0 

-1470.3 

10.14 

7.58 

2923.4 

2185.3 

13 

Pis'  -288.3 

-1.87 

-4.14 

-   539.1 

-1193.6 

5.69 

2.82 

1640.4 

813.0 

15 

Pis'  =  267.  3 

+0.92 

-1.22 

+   245.9 

-   326.0 

-0.10 

-   3.25 

26.7 

-   868.7 

17 

Pn'  =  282.0 

5.86 

+  3.86 

1652.5 

+  1088.5 

-5.50 

-   8.63 

-1551.0 

-2433.7 

19 

Pi9'  =  322.7 

5.86 

5.35 

1891.0 

1726.4 

-9.19 

-12.18 

-2965.6 

-3930.5 

21 

P2i'  =  375.4 

1.98 

1.73 

743.3 

649.4 

-8.55 

-11.39 

-3210.5 

-4275.8 

23 

P24'  =  466.7 

0.33 

0.28 

154.0 

130.7 

-2.15 

-   4.87 

-1003.4 

-2272.8 

7-»  

3508.3 

D- 

6'.  90 

+  1922.5 

-1496.6 

Z>= 

9'.75 

+  1520.5 

-5927.2 

if- 

1485  .  3 

N= 

2291.4 

v  = 

+  OM44 

v= 

-0'.962 

/= 

-   242.3 

-    188.7 

/= 

-     96.0 

-   374.1 

The  influence  line  ordinates  ij  for   points  TO'  and  tf  are  symmetric  with  those   of  points  m  and  t.     See  note,  foot  of 
Table  79i. 


344 


KINETIC  THEORY  OF  ENGINEERING  STRUCTURES          CHAP.  XV 


By  referring  to  Fig.  79E,  it  is  seen  that  this  position  of  loading  gives  maximum  stresses 
for  the  section  I  and  almost  maximum  for  section  m,  but  not  for  the  other  sections.  In 
the  present  instance  the  effects  at  the  five  critical  sections  were  found  for  one  position 
of  the  moving  load,  giving  a  simultaneous  condition  of  stress,  but  not  maximum  stresses 
at  all  sections. 

The  maximum  stresses  for  each  of  the  critical  sections  must  be  computed  for  special 
positions  of  the  moving  train  as  illustrated  in  Table  79K,  for  the  crown  section. 

In  the  general  investigation  of  stresses  it  is  best  to  obtain  the  dead  load  moments 
separately  for  each  section  and  combine  these  with  the  live  load  moments  which  are  found 
without  regard  to  impact  and  then  increased  for  impact  prior  to  adding  the  dead  load 
moments,  as  shown  in  Table  79K. 

TABLE  79K 

INVESTIGATION  FOR  MAXIMUM  STRESSES  AT  THE  CROWN  SECTION  "n." 

SYMMETRIC   ARCH 


Dead 

Total  Dead  Load  Moments. 

No. 

•l-I'L            1 

Max.  +  Live  Load  Moments. 

Point. 

Loads 

vv  neel 

Live 

from 

Loads, 

Q 

ije 

1,1 

Me 

Mi 

Head  of 
Train. 

f)e 

'Ji 

Me 

Mi 

cu.ft. 

ft. 

ft. 

cu.ft. 

ft. 

ft. 

1 

367.4 

-0.14 

-0.18 

-   51.4 

-   66.1 

1 

16.5 

0.00 

-1.19 

0.0 

-   19.6 

3 

276.1 

-0.63 

-6.87 

-173.9 

-240.2 

2 

33.0 

1.30 

+  0.10 

42.9 

+     3.3 

5 

223.4 

-1.19 

-1.78 

-265.8 

-397.6 

3 

33.0 

2.55 

1.10 

84.2 

36.3 

7 

182.7 

-0.71 

-1.70 

-129.7 

-310.6 

4 

33.0 

4.10 

2.55 

135.3 

84.2 

9 

168.0 

+  0.94 

-0.21 

+  157.9 

-  .35.3 

5 

33.0 

6.10 

4.45 

201.3 

146.9 

11 

173.7 

4.92 

+  3.36 

854.6 

+  583.6 

6 

21.4 

6.40 

4.80 

137.0 

102.7 

13 

173.7 

4.92 

+  3.36 

854.6 

+  583.6 

7 

21.4 

4.40 

2.80 

94.2 

59.9 

15 

168.0 

0.94 

-0.21 

157.9 

-  35.3 

8 

21.4 

2.55 

1.00 

54.6 

21.4 

17 

182.7 

-0.71 

-1.70 

-129.7 

-310.6 

9 

21.4 

1.30 

0.00 

27.8 

0.0 

19 

223.4 

-1.19 

-1.78 

-265.8 

-397.6 

10 

0.0 

0.00 

-1.19 

0.0 

0.0 

21 

27fi   1 

—  0  fi^ 

087 

mo 

74O   9 

ft  A 

23 

•fff  \f  •  A 

367.4 

\J  .  UO 

-0.14 

.  01 

-0.18 

.   U 

-   51.4 

—  HI  .  £, 

-  66.1 

Max.+    L.L.= 

+  777.3 

+  435.1 

84% 

652.9 

+  365.5 

R  = 

2782.6 

Z>  = 

6'.  00 

+  783.4 

-932.4 

N  = 

857.9 

TotaFD.L.= 

783.4 

-932.4 

v  = 

-0'.087 

Total  max.  +  A/  = 

2213.6 

-131.8 

f= 

-130.6 

-155.4 

max.A'r  = 

1190.8 

»=-<y.88 

max./= 

-369.0 

-   22.9 

1 

All  stresses  /  are  in  cubic  feet  per  square  fook     Multiply  by  0.972  to  reduce  to  pounds  per  square 

300 

inch.     Impact  for  56  feet  is  —        —=0.84. 
oUO+56 


The  live  load  moments  are  found  by  placing  the  train  of  wheel  loads  for  Cooper's 
E  60  loading  over  the  positive  portions  of  the  Me  and  M.,  influence  lines  for  the  section 
in  question  to  obtain  the  simultaneous  maximum  positive  moments  for  the  kernel  points. 
Similarly  the  negative  portions  of  these  influence  lines  will  yield  the  maximum  negative 
moments  for  the  same  section. 


ART.  79  FIXED  MASONRY  ARCHES  345 

Since  the  loads  are  distributed  over  a  roadbed  13  feet  in  width  the  train  of  wheel 
loads,  for  the  present  purpose,  is  obtained  by  dividing  the  standard  axle  loads  by  13,  and 
also  converting  the  loads  into  cubic  feet  of  masonry,  all  as  shown  in  Fig.  79o.  The  load 
diagram  must  of  course  be  drawn  to  the  same  horizontal  scale  as  the  influence  lines,  which 
is  not  the  case  in  the  figures  as  here  reproduced. 

The  wheel  loads,  in  kips,  are  converted  into  cubic  feet  of  masonry  by  multiplying 
kips  by  1000/140-7.14. 

The  maximum  stresses  at  the  crown  section  n  are  computed  in  Table  79ic.  The  dead 
load  moments  are  first  found  and  these  are  combined  with  the  live  load  moments  which 
latter  are  increased  for  impact.  The  maximum  stresses  are  found  from  the  total  maximum 
moments. 

The  live  load  diagram  Fig.  79c,  was  placed  over  the  influence  line  for  section  n,  Fig. 
70E,  so  that  the  first  load  came  at  the  point  8,  which  is  the  zero  point  for  positive  ordinates. 
One  locomotive  then  covered  the  distance  from  point  8  to  16,  and  the  remainder  of  the 
span  was  considered  empty.  The  same  position  was  used  for  both  Me  and  Mi  moments, 
because  Eqs.  (76c)  apply  only  to  simultaneous  kernel  moments. 


UNIFORM  J.OAO  OP  0.461  kiM  p.tT. 


V       $'     5'     S' 

FIG.  79c. — Cooper's  E  60  Loading  for  an  Arch  Ring  One  Foot  Thick. 

Comparing  the  result  for  max.  fe=  —369  cu.ft.,  obtained  here  with  the  value  given 
in  Table  79j  as— 290  cu.ft.,  the  relative  effect  of  the  two  cases  of  loading  is  clearly  seen. 

Each  of  the  critical  sections  should  be  examined  in  the  manner  shown  in  Table 
79K,  and  the  positive  as  well  as  negative  influence  line  areas  should  be  considered  as  sepa- 
rate cases. 

The  temperature  stresses  are  easily  determined  according  to  the  method  given  in 
Art.  75. 

Referring  back  to  the  given  data,  E  =3,670,000  Ibs.  per  sq.in.  =3,755,720  cu.ft. 
per  sq.ft.;  £=0.000,005,4  per  1°  F;  £=-80°;  Z  =  158.8  ft;  and  cos  /?  =  !,  then  etf  = 
-0.0686  ft.,  and  by  Eq.  (75c) 

dlE  0.0686  X3,775,720 


where  the  value  cos2  /?£t/TFc  is  taken  from  Table  79c  for  Wc  loads  which  are  E  times 
actual. 

The  stresses  on  any  section  are  then  found  by  evaluating  the  kernel  moments  for 
temperature  and  combining  these  with  the  kernel  moments  previously  found  for  the  case 
of  maximum  loading. 


346  KINETIC  THEORY  OF  ENGINEERING  STRUCTURES          CHAP.  XV 

Thus,  for  the  haunch  section  t,  Table  79o  gives  D  =9'. 75,  ye  =  -37'. 1  and  t/t-=  -35'.3 
and  maximum  Me  and  Mi  are  taken  from  Table  79j  as  follows: 

Met=  -37.1  X  230.9=  -8566.4 
Me     from  Table  79 j-  +6973.9 

Total  Me  =-1592.5 

....        ,  6X1592.5       ,  inn  „        ,. 

giving  fe  =  + ,     =  +  lOO.o  cu.ft. 

9.7o~ 

Also, 

Mil=  -35.3X230.9=  -8150.8 
Mi    from  Table  79 j  =  +  453.2 

Total  Mt  =-7697. 6 

.  .      ,         6X7697.6         ,.e          ,, 

giving/,  = .,    =  —458.8  cu.ft. 

9.75" 

Similarly  for  the  crown  section  n  where  Z>=6'.0,  ye  =  ll'.OS  and  i/;  =  13'.08,  then  for 
the  maximum  moments  from  Table  79K 

Met=  10.08  X  230.9  =  +  2558.4 
Me  from  Table  79K -  +  2213.6 

Total  Me=  +4772.0  or/e  =  -795.3  cu.ft., 

and  Mit  =  13.08  x  230.9  =  +  3020.2 

Mi  from  Table  79K  =  -    131.8 


Total  M;=  +2888.4  or /»•=  +481.4  cu.ft. 

It  is  readily  seen  that  these  stresses  are  excessive  and  would  undoubtedly  cause 
cracks  on  the  tension  elements  occurring  in  the  intrados  at  the  crown  and  in  the  extrados 
at  the  haunches. 

(c)  Analysis  as  under  (b),  treating  the  structure  as  unsymmetric.  It  is  presumed 
that  for  good  and  sufficient  reasons  the  B  abutment  had  to  be  founded  on  a  lower  stratum 
of  rock,  a  condition  which  developed  after  the  original  design  for  the  symmetric  arch 
was  accepted.  The  question  to  be  answered  here  is  whether  such  a  change  in  the  location 
of  a  foundation  would  seriously  alter  the  resulting  stresses  and  necessitate  a  change  in 
the  design. 

The  location  of  the  (x,  y)  axes  and  computation  of  the  elastic  loads  is  given  in  Tables 
79L  and  79.M,  following  the  method  outlined  in  Art.  73,  and  using  the  same  data  as  in  the 
previous  case  of  the  symmetric  arch  with  the  addition  of  two  sections  at  the  right-hand 
end,  producing  the  unsymmetric  condition  illustrated  in  Fig.  79u. . 

The  (z,  v)  axes  are  located  through  the  axial  points  of  the  haunches,  which  is  the 
same  position  previously  chosen  for  these  axes  in  dealing  with  the  symmetric  condition. 

The  dimensions  D,  v  and  z  are  tabulated  in  Table  79L,  and  from  these  the  values 
1/7,  sec  (j>,  wa,  vwa  and  zwa  are  computed,  using  the  formulae  for  sec  <£  above  given,  p.  331. 


ART.  79 


FIXED  MASONRY  ARCHES 


347 


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348 


KINETIC  THEORY  OF  ENGINEERING  STRUCTURES 


TABLE 
COMPUTATION   OF   W  FORCES   AND   POLE 


Pt. 

Wa 

XZWa 

.V. 

=  xz  Wa 

^ 

Eq.  (73c), 

?/  —  2  —  2  '  ~\~  X  T  '  ill  3 

XWa 

x  tan  3 

2  —  20' 

y 

ft. 

0 

0.0232 

49.8 

153.2 

155.1 

446.6 

-4.22 

-37.17 

-41.39 

1.885 

1 

0.0251 

66.6 

142.6 

414.5 

885.0 

-3.91 

-28.37 

-32.28 

1.892 

2 

0.0285 

85.0 

135.2 

551.2 

877.8 

-3.57 

-20.27 

-23.84 

1.963 

3 

0.0325 

101.4 

126.4 

656.0 

817.5 

-3.24 

-13.57 

-16.81 

2.027 

4 

0.0354 

110.4 

110.5 

716.8 

718.8 

-2.90 

-  7.77 

-10.67 

1.978 

5 

0.0394 

118.3 

96.0 

765.2 

622.3 

-2.56 

-  2.77 

-  5.33 

1.945 

6 

0.0429 

119.7 

78.8 

774.2 

511.5 

-2.22 

+  1.53 

-  0.69 

1.839 

7 

0.0458 

114.4 

60.6 

739.5 

393  .  9 

-1.89 

5.13 

+  3.24 

1.666 

8 

0.0477 

102.0 

42.6 

661.0 

278.3 

-1.55 

8.03 

6.48 

1.425 

9 

0.0499 

86.1 

27.3 

556.8 

178.8 

-1.21 

10.23 

9.02 

1.166 

10 

0.0513 

65.2 

14.6 

450.6 

102.8 

-0.88 

11.73 

10.85 

0.865 

11 

0.0530 

38.0 

4.6 

283.2 

38.0 

-0.49 

12.93 

12.44 

0.497 

12 

0.0555 

8.0 

0.2 

51.2 

5.2 

-0.10 

13.23 

13.13 

0.104 

13 

0.0530 

-  22.8 

1.7 

-168.6 

16.3 

+  0.29 

12.93 

13  .  22 

-0.298 

14 

0.0513 

-  50.7 

8.8 

-349.1 

62.8 

0.68 

11.73 

12.41 

-0.674 

15 

0.0499 

-  72.3 

19.2 

-467.3 

126.4 

1.02 

10.23 

11.25 

-0.980 

16 

0.0477 

-  89.2 

32.6 

-577.8 

213.4 

1.36 

8.03 

9.39 

-  1  .  246 

17 

0.0458 

-102.7 

48.8 

-663.6 

317.5 

1.69 

5.13 

6.82 

-1.494 

18 

0.0429 

-109.3 

65.7 

-706.6 

426.6 

2.03 

1.53 

3.56 

-  1  .  679 

19 

0.0394 

-109.3 

82.0 

-706.8 

531.7 

2.37 

-  2.77 

-  0.40 

-  1  .  798 

20 

0.0354 

-103.0 

96.2 

-668.5 

625.8 

2.71 

-  7.77 

-  5.06 

-1.845 

21 

0.0325 

-  95.3 

111.7 

-615.4 

722.4 

3.04 

-13.57 

-10.53 

-1.906 

22 

0.0285 

-  80.4 

120.9 

-521.2 

784.9 

3.38 

-20.27 

-16.89 

-  1  .  856 

23 

0.0251 

-  63.3 

128.8 

-388.0 

782.1 

3.72 

-28.37 

-24.65 

-  1  .  798 

24 

0.0174 

-  35.6 

104.6 

-214.3 

621.6 

4.02 

-37.17 

-33.15 

-1.349 

25 

0.0129 

-  16.7 

89.8 

-  99.6 

544.1 

4.33 

-48.07 

-44.74 

-1.076 

26 

0.0131 

0.0 

104.5 

-  8.2 

301.0 

4.64 

-63.57 

-58.93 

-1.170 

+  620.3 

11953.1 

All  elastic  loads  W 

The  sums  ^wa  =  Wa,  ^vwa=vWa  and  Hzwa=zWa  are  now  computed  by  Simpson's 
rule  according  to  Eqs.  (73E),  using  the  lengths  Jv  for  the  horizontal  distances  between 
the  arch  sections. 

Eqs.  (73r)  then  give  the  coordinates  l\,  z0'  of  the  new  origin  0  with  respect  to  the 
(v,  z)  axes.  Thus 

2vWa     545.7264 


6.7149 


=81.272  feet, 
=63.574  feet, 


and  x=li  —v. 


ART.  79 


FIXED  MASONRY  ARCHES 


349 


79M 

DISTANCES.     UNSYMMETRTC  ARCH 


ft. 

z^-r, 

ywa 

2  yWa  —  We 

y2 

* 

» 

cos  <f> 

2  cos  <j>Jx 

Point. 

D 

D 

5.56 

-0.960 

-  2.76 

1713.1 

39.74 

110.5 

0  052 

0.158 

0 

5.9 

11.77 

-0.810 

-  5.04 

1042.0 

26.15 

163  .  4 

0.064 

0.394 

1 

6.5 

12.77 

-0.679 

-  4.43 

568.3 

16.20 

106  8 

0.077 

0.500 

2 

6.5 

13.13 

-0.546 

-  3.54 

282  6 

9.18 

60.6 

0  090 

0.583 

3 

6.5 

12.88 

-0.378 

-  2.47 

113.8 

4.03 

27.4 

0.101 

0.657 

4 

6.5 

12.61 

-0.210 

-  1.36 

28.4 

1.12 

8.3 

0.113 

0.732 

5 

6.5 

11.92 

-0.030 

-  0.13 

0.5 

0.02 

1.0 

0.123 

0.799 

6 

6.5 

10.80 

+  0.148 

+  0.95 

10.5 

0.48 

3.7 

0.135 

0.872 

7 

6.5 

9.26 

0.309 

1.99 

42.0 

2  00 

13.3 

0.144 

0.930 

8 

6.5 

7.56 

0.450 

2.90 

81.4 

4.06 

26.3 

0.153 

0.987 

9 

6.5 

5.99 

0.557 

3.89 

117.7 

6.04 

42.6 

0.158 

1.110 

10 

7.5 

3.71 

0.659 

4.91 

154.8 

8.20 

61.0 

0.163 

1.225 

11 

7  .5 

0.73 

0.729 

5.39 

172.4 

9.57 

70.7 

0.167 

1.250 

12 

7.5 

-  2.22 

0.701 

5.22 

174.8 

9.26 

68.8 

0.163 

1.225 

13 

7.5 

4.65 

0.637 

4.45 

154.0 

7.90 

55.3 

0.158 

1.110 

14 

6.5 

6.35 

0.561 

3.62 

126.6 

6.32 

40.8 

0.153 

0.987 

15 

6.5 

-  8.08 

0.448 

2.89 

88.2 

4.21 

27.4 

0.144 

0.930 

16 

6.5 

-  9.66 

0.312 

2.01 

46.5 

2.13 

-  14.1 

0.135 

0.872 

17 

6.5 

-  10.86 

0.153 

0.99 

12.7 

0.54 

4.1 

0.123 

0.799 

18 

6.5 

-  11.63 

-0.016 

-  0.08 

0.2 

0.01 

0.8 

0.113 

0.732 

19 

6.5 

-  11.98 

-0.179 

-  1.16 

25.6 

0.91 

6.9 

0.101 

0.657 

20 

6.5 

-  12.31 

-0.342 

-  2.21 

110.9 

3.60 

24.4 

0.090 

0.583 

21 

6.5 

-  12.05 

-0.481 

-  3.14 

285.3 

8.13 

54.2 

0.077 

0.500 

22 

6.5 

-  10.94 

-0.619 

-  3.75 

607.6 

15.25 

92.6 

0.064 

0.393 

23 

5.9 

8.03 

-0.577 

-  3.44 

1098.9 

19.12 

114.2 

0.047 

0.275 

24 

5.9 

6.53 

-0.577 

-3.52 

2001  .  7 

25.82 

158.7 

0.032 

0.187 

25 

5.9 

3.40 

-0.772 

-  2.18 

3472.7 

45.49 

124.5 

0.020 

0.064 

26 

+  118.69 

+  39.21 

1482.4 

19.511 

-118.69 

-39.21 

«.#:,*. 

=  1480.6 

are  E  times  actual. 

When  the  ordinates  x  are  found  then  the  quantities  xzwa  and  x2wa  and  their  sums 
xzWa  and  xWi,  are  determined  in  Table  79M,  using  Simpson's  rule  in  summing  for  the 
horizontal  increments  Ax. 

Eq.  (73n)  then  gives 

620.3 


tan  8  = 


11953.1 


=  -0.0519. 


Eq.  (73o)  now  furnishes  the  ordinates  y  and  from  these  the  quantities  ywa  and 
y2wa  are  computed.  As  a  check  STF&=0  and  SWC=0,  and  any  small  discrepancies 
must  be  adjusted. 


350 


KINETIC  THEORY  OF  ENGINEERING  STRUCTURES          CHAP.  XV 


ART.  79 


FIXED  MASONRY  ARCHES 


351 


The  W  loads  were  again  derived  without  including  E  so  that  these  loads  are  all  E 
times  actual,  a  fact  which  should  not  be  overlooked  if  temperature  stresses  are  to  be 
found  from  Eq.  (75c). 

In  Table  79M  the  quantities  cos  (f>dx/D  are  given,  showing  that  the  pole  distance 
Hc=cos  52?/TFc  +  2cos  (f>Ax/D  from  Eq.  (72ic),  would  be  increased  by  1.3  per  cent  if  axial 
thrust  is  included  in  the  determination  of  Xc.  This  would  tend  to  diminish  Xc,  and  the 
ordinates  of  the  Xc  influence  line. 

The  influence  lines  for  Xa,  Xb  and  Xc  are  shown  in  Fig.  79i,  noting  that  the  length  of 
span  /,  is  now  taken  as  the  horizontal  distance  between  the  intrados  kernel  points  of  the 
haunch  sections.  This  was  done  to  obviate  the  negative  M0  moments  otherwise  encoun- 
tered here  and  leading  to  complications  which  are  rather  far  reaching.  In  the  previous 
case  of  symmetric  arch,  the  real  haunch  sections  could  not  be  examined  becuase  the  span 
was  chosen  as  the  distance  between  the  verticals  through  the  axial  points  and  the  sec- 
tion t  was  accordingly  taken  as  the  nearest  approach  to  the  haunch  section.  It  is  thus 
interesting  to  note  the  two  methods  of  treating  this  matter.  The  length  does  not 
enter  into  the  computations,  but  must  be  considered  in  drawing  all  the  influence 
lines. 

The  detailed  description  of  the  method  of  drawing  these  influence  lines  need  riot 
be  repeated  here,  and  the  reader  is  referred  back  to  the  treatment  of  the  symmetric  arch 
in  the  first  part  of  this  article. 

Axial  thrust  was  not  included  in  Fig.  79i,  thus  making  all  the  results  of  both  investiga- 
tions comparable.  However,  the  effect  is  so  small  that  it  may  be  regarded  as  a  negligible 
quantity  in  the  present  case. 

Having  located  the  coordinate  axes,  this  determines  the  critical  sections,  and  the  data 
required  for  the  computation  of  the  ordinates  for  the  Me  and  Mi  influence  lines  are  given 
in  Table  79N. 


TABLE    79N.— DATA    RELATING    TO    THE    CRITICAL    SECTIONS— UNSYMMETRIC    ARCH 


Coordinates, 

Coordinates  Extrados  Kernel 

Coordinates  Intrados  Kernel 

Axial  Point. 

Points. 

Points. 

T>     ' 

D    . 

D        , 

X 

y 

g  sm  <j> 

g    COS  $ 

xe 

Ve 

cos  @ye 

*i 

Vi 

COS  &/j 

ft. 

ft. 

ft. 

ft. 

ft. 

ft. 

ft. 

ft. 

ft. 

ft. 

ft. 

A 

10.0 

58°  51' 

+  81.27 

-41.39 

1.43 

0.86 

+  79  .  85 

-42.25 

-42.20 

+  82.70 

-40.53 

-40.48 

m 

6.9 

31     04 

+  41.20 

0.00 

0.59 

1.00 

+  40.61 

-    1.00 

-    1.00 

+  41.79 

+    1.00 

+    1.00 

n 

6.0 

0    00 

0.00 

+  13.25 

0.00 

1.00 

0.00 

+  12.25 

+  12.23 

0.00 

+  14.25 

+  14.23 

m" 

7.2 

34    58 

-43.90 

0.00 

0.69 

0.98 

-43.21 

-  0.98 

-   0.98 

-44.59 

+  0.98 

+   0.98 

B 

14.7 

73     11 

-89.33 

-58.93 

2.35 

0.71 

-86.98 

-59.64 

-59.56 

-91.68 

-58.22 

-58.14 

/?=2°  58'  and  cos  0  =  0.9987  from  Table  79L 

The  M0  influence  lines  for  the  kernel  points  of  the  several  critical  sections  are  shown 
in  Fig.  79J,  the  data  for  the  construction  of  these  being  taken  from  Table  79x.  It  should 
be  noted  that  for  the  haunch  sections  A  and  B,  Mot-=0. 


352 


KINETIC  THEORY  OF  ENGINEERING  STRUCTURES          CHAP.  XV 


1=174138 


FIG.  79i. — X  Influence  Lines  for  Unsymmetric  Arch 


ART.  79 


FIXED  MASONRY  ARCHES 


353 


From  the  coordinates  of  the  kernel  points  in  Table  79N,  and  the  ordinates  in  Tables 
79o  and  79p,  taken  from  the  Xa,  Xb,  Xc  and  M0  influence  lines,  Figs.  79i  and  79J,  the  T)e 
and  j)i  are  computed  for  the  odd  points  and  for  the  five  critical  sections  according  to  Eqs. 
(76B).  The  operations  are  all  indicated  in  the  headings  of  Tables  79o  and  79p,  thus 
requiring  no  further  description  here. 

These  ordinates,  plotted  in  Fig.  79K,  furnish  the  Me  and  M{  influence  lines  from 
which  the  moments  and  stresses  for  any  position  of  the  live  loads  may  be  found  for  each 
of  the  five  critical  sections.  In  practice  it  is  well  to  compute  all  the  ordinates  TJ 
instead  of  merely  the  odd  ones  as  given  here. 

FOR  INTRADOS  KERNEL  POINTS. 


FORCXTRAOOS  KERNEL  POINTS, 


FIG.  79j. — M0  Influence  Lines  for  Unsymmetric  Arch. 


Table  79o,  gives  the  computations  for  the  resultant  polygon  for  the  same  case  of 
loading  previously  used  in  Table  79n,  adding  one  additional  load  P^5  at  a  point  on  the 
right  abutment.  The  results  of  this  computation  are  shown  in  Fig.  79n,  where  the  resultant 
polygon  is  drawn  through  the  points  a  and  b  which  are  located  on  the  verticals  through 
the  intrados  kernel  points  of  the  A  and  B  sections. 

It  should  be  noted  that  c  and  c0  are  both  negative,  indicating  that  they  must  be  meas- 
ured up  from  the  line  ah  when  found.  Hence  z0  is  measured  down  on  the  y  axis  to 
locate  the  point  s  and  —  c0=ss'  is  then  laid  off  upward  from  s.  The  line  as'||z 
axis,  locates  the  point  a  at  the  left  abutment  and  the  line  as  prolonged  fixes  the  point  6 
at  the  right  abutment.  The  force  polygon  is  therl  easily  constructed  by  laying  off 
the  reactions  A0  and  B0  and  finding  the  pole  0\  by  drawing  the  line  0\T  \\  ab  and  equal  in 
length  to  H', 


354 


KINETIC  THEORY  OF  ENGINEERING  STRUCTURES          CHAP.  XV 

TABLE   79o.— MOMENT  INFLUENCE   LINES   FOR 


Point. 

90 

ft. 

fib 
ft. 

rye 
ft. 

SECTION  A. 

r>  A  =  r>  -  [ij  a+  79.857/6  -  42.27jc] 

SECTION  TO. 
rjm  =  ijo  —  [i)a  +  40.6j?&  —  l.Oqc] 

rlo 

79.85  t)b 

-42.2r,c 

IA 

T>0 

+  40.67,6 

T)m 

1 

3.5 

0.036 

0.012 

2.74 

2.87 

-  0.51 

-   3.12 

5.2 

1.46 

0.2t> 

3 

9.8 

0.082 

0.100 

2.52 

6.55 

-   4.22 

-   9.61 

15.2 

3.33 

2.17 

5 

15.2 

0.102 

0.252 

2.31 

8.14 

-10.63 

-10.40 

25.1 

4.14 

6.01 

7 

19.7 

0.092 

0.424 

2.10 

7.35 

-17.89 

-   7.06 

31.0 

3.74 

7.98 

9 

22.9 

0.064 

0.580 

1.88 

5.11 

-24.48 

-    1.65 

28.0 

2.60 

3.08 

11 

25.0 

0.014 

0.692 

1.64 

1.12 

-29.20 

+  4.72 

24.5 

0.57 

-0.38 

13 

25.5 

-0.044 

0.720 

1.40 

-   3.51 

-30.38 

9.79 

21.0 

-1.79 

-1.99 

15 

24.5 

-0.100 

0.652 

1.18 

-   7.99 

-27.51 

12.18 

17.4 

-4.06 

-2.39 

17 

22.1 

-0.132 

0.520 

0.97 

-10.54 

-21.94 

11.35 

14.2 

-5.36 

-2.02 

19 

18.7 

-0.148 

0.360 

0.75 

-11.62 

-15.19 

8.86 

11.1 

-6.01 

-1.23 

21 

14.3 

-0.136 

0.196 

0.53 

-10.86 

-   8.27 

5.36 

8.0 

-5.52 

-0.58 

23 

9.1 

-0.100 

0.070 

0.32 

-   7.99 

-   2.95 

2.16 

4.8 

-4.06 

-0.17 

25 

3.9 

-0.046 

0.016 

0.13 

-   3.67 

-   0.68 

0.58 

2.0 

-1.87 

-0.01 

All  ordinates  in  above 


TABLE    79p.— MOMENT   INFLUENCE    LINES   FOR 


Point. 

T/a 
ft. 

>?& 
ft. 

T)c 
ft. 

SECTION  A. 
^  =  0-[7ja+82.77j6-40.5rjc] 

SECTION  m. 

•fjm  =  rjo  —  [f)a  +  41  .87J6  +  1.0)jf] 

S2.7rjb 

-40.57yc 

r>A 

1)0 

41.87/6 

l.OlJc 

TJm 

1 

3.5 

0.036 

0.012 

2.98 

-   0.49 

-   5.99 

5.4 

1.50 

0.01 

0.39 

3 

9.8 

0.082 

0.100 

6.78 

-  4.05 

-12.53 

15.4 

3.43 

0.10 

2.07 

5 

15.2 

0.102 

0.252 

8.44 

-10.21 

-13.43 

25.2 

4.26 

0.25 

5.49 

7 

19.7 

0.092 

0.424 

7.61 

-17.17 

-10.14 

30.0 

3.85 

0.42 

6.03 

9 

22.9 

0.064 

0.580 

5.29 

-23.49 

-   4.70 

27.0 

2.68 

0.58 

0.84 

11 

25.0 

0.014 

0.692 

1.16 

-28.03 

+    1.87 

23.8 

0.59 

0.69 

-2.48 

13 

25.5 

-0.044 

0.720 

-   3.64 

-29.16 

7.30 

20.2 

-1.84 

0.72 

-4.18 

15 

24.5 

-0.100 

0.652 

-   8.27 

-26.41 

10.18 

17.0 

-4.18 

0.65 

-3.97 

17 

22.1 

-0.132 

0.520 

-10.92 

-21.06 

9.88 

14.0 

-5.52 

0.52 

-3.10 

19 

18.7 

-0.148 

0.360 

-12.24 

-14.58 

8.12 

11.0 

-6.18 

0.36 

-1.88 

21 

14.3 

-0.136 

0.196 

-11.25 

-   7.94 

4.89 

7.8 

-5.68 

0.20 

-1.02 

23 

9.1 

-0.100 

0.070 

-  8.27 

-   2.84 

2.01 

4.8 

-4.18 

0.07 

-0.19 

25 

3.9 

-0.046 

•0.016 

-   3.80 

-   0.65 

0.55 

2.0 

-1.92 

0.02 

0.00 

All  ordinates  in  above 

It  is  seen  that  the  resultant  polygon  intersects  the  axial  line  four  times,  showing  con- 
sistency with  Professor  Winkler's  theorem. 

As  a  final  check,  the  normal  thrusts  TV,  offsets  v  and  stresses /are  computed  in  Table 
79n  for  the  same  case  of  loading;  used  in  Table  79Q,  and  the  close  agreement  of  the  results 
shows  both  solutions  to  be  satisfactory.  The  only  stresses,  however,  which  are  maximum, 
are  those  on  the  section  A,  the  others  are  simply  for  the  simultaneous  case  of  loading 
and  have  no  special  interest. 

The  stresses  on  the  haunch  section  A  are  thus  found  to  be/e=  —549  and/t-=  +152.9 


ART.  79 


FIXED  MASONRY  ARCHES 


355 


EXTRADOS   KERNEL  POINTS— UNSYMMETRIC  ARCH 


SECTION  n. 

,„  =  ,„-  [,a+12.23,c] 

SECTION  m'. 

SECTION  B. 

Point. 

r/o 

12.23  7jc 

r/n 

VO 

-43.2,6 

-.098,c 

T)m 

fjo 

-87.0,6 

-59.56,c 

r,B 

3.7 

0.14 

+  0.05 

1.9 

-1.56 

-0.01 

-0.03 

0.18 

-   3.13 

-  0.71 

0.52 

1 

10.6 

1.22 

-0.42 

5.5 

-3.54 

-0.10 

-0.66 

0.54 

-   7.13 

-   5.96 

3.83 

3 

17.2 

3.08 

-1.08 

9.1 

-4.41 

-0.25 

-1.44 

0.89 

-  8.88 

-15.01 

9.58 

5 

24.2 

5.19 

-0.69 

13.0 

-3.97 

-0.41 

-2.32 

1.24 

-  8.00 

-25.25 

14.79 

7 

31.1 

7.10 

+  1.10 

16.4 

-2.76 

-0.57 

-3.17 

1.59 

-   5.57 

-34.54 

18.80 

9 

38.5 

8.46 

5.04 

20.3 

-0.60 

-0.68 

-3.42 

1.99 

-    1.22 

-41.22 

19.43 

11 

40.8 

8.80 

6.50 

24.6 

+  1.90 

-0.71 

-2.09 

2.38 

+   3.83 

-42.88 

15.93 

13 

34.2 

7.97 

+  1.73 

28.4 

4.32 

-0.64 

+  0.22 

2.76 

8.70 

-38.83 

8.39 

15 

28.1 

6.36 

-0.36 

32.0 

5.70 

-0.51 

4.71 

3.10 

11.48 

-30.97 

0.49 

17 

22.0 

4.40 

-1.10 

33.4 

6.39 

-0.35 

8.66 

3.47 

12.88 

-21.44 

-   6.67 

19 

15.6 

2.40 

-1.10 

23.7 

5.88 

-0.20 

3.72 

3.82 

11.83 

-11.67 

-10.64 

21 

9.5 

0.85 

-0.45 

14.4 

4.32 

-0.07 

+  1.05 

4.17 

8.70 

-  4.17 

-  9.46 

23 

4.0 

0.20 

-0.10 

5.8 

1.99 

-0.02 

-0.07 

4.50 

4.00 

-   0.95 

-   2.45 

25 

table  are  expressed  in  feet. 


INTRADOS   KERNEL   POINTS— UNSYMMETRIC    ARCH 


SECTION  n. 
'Jn=  >?o  —  [)Ja+  14.23^c] 

SECTION  m'. 
5}m/==)?o  -[Jjo  —  44.6i;6  +  0.987jc] 

SECTION  B. 
?B  =  0-[?«-9l.7ij6-58.14ijc] 

Point. 

rio 

14.23ijc 

Tjn 

90 

-44.67?6 

0.98)?c 

Tim 

-91.7>j6 

-58.14/jc 

IB 

3.7 

0.17 

+  0.03 

1.8 

-1.61 

0.01 

-0.09 

-3.30 

-   0.70 

+   0.50 

1 

10.6 

1.42 

-0.62 

5.4 

-3.66 

0.10 

-0.84 

-7.52 

-   5.81 

3.53 

3 

17.2 

3.58 

-1.58 

8.9 

-4.55 

0.25 

-2.00 

-9.35 

-14.65 

8.80 

5 

24.2 

6.03 

-  1  .  53 

12.4 

-4.10 

0.41 

-3.61 

-8.44 

-24.65 

13.39 

7 

31.1 

8.25 

-0.05 

16.0 

-2.85 

0.57 

-4.62 

-5.87 

-33.72 

16.69 

9 

38.5 

9.85 

+  3.65 

19.6 

-0.62 

0.68 

-5.46 

-1.28 

-40.23 

16.51 

11 

40.8 

10.24 

5.06 

23.8 

+  1.96 

0.71 

-4.37 

+  4.03 

-41.86 

12.33 

13 

34.2 

9.28 

0.42 

27.8 

4.46 

0.64 

-1.80 

9.17 

-37.91 

4.24 

15 

28.1 

7.40 

-1.40 

31.2 

5.88 

0.51 

+  2.71 

12.10 

-30.23 

-  3.97 

17 

22.0 

5.12 

-1.82 

33.4 

6.60 

0.35 

7.75 

13.57 

-20.93 

-11.34 

19 

15.6 

2.84 

-1.54 

24.2 

6.06 

0.20 

3.64 

12.47 

-11.40 

-15.37 

21 

9.5 

0.99 

-0.59 

14.6 

4.46 

0.07 

0.97 

9.17 

-  4.07 

-14.20 

23 

4.0 

0.23 

-0.13 

6.0 

2.05 

0.02 

0.03 

4.22 

-   0.93 

-  7.19 

25 

table  are  expressed  in  feet. 

cu.ft.,  which  are  quite  excessive  as  compared  to  those  previously  obtained  for   section 
t  of  the  symmetric  arch,  where  fe  =  —440  and  J\  =  +28.5  cu.ft. 

If  the  investigation  of  stresses  were  extended  to  all  the  critical  sections  it  would 
be  found  that  the  dimensions  found  safe  for  the  symmetric  arch  are  now  quite  insufficient 
when  the  arch  is  treated  as  an  unsymmetric  structure  using  the  same  axial  line  as  before. 
The  shape  of  the  ring  might  be  so  adjusted  as  to  reduce  the  stresses  within  the  required 
limits  by  computing  a  resultant  polygon  for  the  average  loading  Q+%P,  and  shifting 
the  axial  line  by  the  amounts  ±v  obtained  for  the  several  critical  sections. 


356 


KINETIC  THEORY  OF  ENGINEERING  STRUCTURES         CHAP.  XV 


A 


^^rrr^r^        SECTION  n.  ^r^rrri' 

•rfrl'"-T        T^jNa   ip    n    la    is    K  i5/fcJ"'T   T— 

m      5*7     8S   X    ]       T  /    T>I7      18    19    ; 


V 

intrados  kernel  points.    \ 


—  — extrados  ke 


rnel  points.      \  //uml 
\  /    '~~^ 


/  LENGTHS. 

/       l...iliml  I  I  I  I  l| 

/       10         O         |O       ZO       3O       4O       SO  PT. 


OROINATCS. 


III! 


SFT. 


v' 
FIG.  79K. — M  Influence  Lines  for  the  Kernel  Points.    Unsymmetric  Arch. 


ART.  79 


FIXED  MASONRY  ARCHES 


357 


TABLE  79q. 
THE  RESULTANT  POLYGON  FOR  LOADS   Q  +  P— UNSYMMETRIC   ARCH 


Dead 

Live 

Total 

r,(Q  +  P)  for 

Moments  about  a 

Point. 

ria 

16 

r,c 

Loads. 

Loads. 

Loads. 

Q 

P 

xa 

xb 

Xc 

r 

r(Q+P)' 

ft. 

ft. 

ft. 

cu.ft. 

cu.ft. 

cu.ft. 

ft. 

1 

3.5 

0.036 

0.012 

367.4 

0.0 

367.4 

1285.9 

13.23 

4.40 

5.9 

2167.7 

3 

9.8 

0.082 

0.100 

276.1 

0.0 

276.1 

2705.8 

22.64 

27.61 

18.9 

5218.3 

5 

15.2 

0.102 

0.252 

223  .4 

0.0 

223.4 

3395.7 

22.82 

56.30 

31.9 

7126.5 

7 

19.7 

0.092 

0.424 

182.7 

0.0 

182.7 

3599.2 

16.81 

77.46 

44.9 

8203.2 

9 

22.9 

0.064 

0.580 

168.0 

0.0 

168.0 

3847.2 

10.75 

97.44 

57.9 

9727.2 

11 

25.0 

0.014 

0.692 

173.7 

114.6 

288.3 

7207  .  5 

4.03 

199  .  50 

71.9 

20728.8 

13 

25.5 

-0.044 

0.720 

173.7 

114.6 

288.3 

7344.8 

-12.68 

207.59 

86.9 

25053.3 

15 

24.5 

-0.100 

0.652 

168.0 

99.3 

267.3 

6548.9 

-26.73 

174.28 

100.9 

26970.6 

17 

22.1 

-0.132 

0.520 

182.7 

99.3 

282.0 

6232.2 

-37.22 

146.64 

113.9 

32119.8 

19 

18.7 

-0.148 

0.360 

223.4 

99.3 

322.7 

6034.5 

-47.76 

116.17 

126.9 

40950.6 

21 

14.3 

-0.136 

0.196 

276.1 

99.3 

375.4 

5368.2 

-51.05 

73.58 

139.9 

52518.5 

23 

9.1 

-0.100 

0.070 

367.4 

99.3 

466.7 

4247.0 

-46.67 

32.67 

152.9 

71358.4 

25£ 

3.5 

-0.040 

0.012 

680.0 

84.0 

764.0 

2674.0 

-30.56 

9.17 

165.9 

126747.6 

4272.3 

60490.9 

-162.39 

1222.81 

428890.5 

=  # 

=  Xa 

=xb 

=XC 

=  M 

Z  =  174'.38,       r,  =  82'.7,       / 
Then  H  =  Xc  cos  /?=  1221.22  cu.ft, 

TT 

H'  = =  1242  cu.ft. 

cos  a 

I 

c  =  -X&=-23.19ft. 
n 

M 

r  =  7r=  100.388  ft. 
ti 


tan  13=  -0.0519,       cos/3  =  0.9987 

V 


tan  a  =  tan 


-77 

£1 


=  -0.1849 


=       =  49.53  ft. 


-^Xi--  11.00  ft. 

H 


B0= 


=2494.6  cu.ft. 


A0  =  R  —  B0  =  1777  7  cu.ft. 

The  investigation  for  temperature  is  not  repeated  here  except  to  evaluate  the  moment 
Eq.  (75c)  for  this  case  using  the  same  data  just  given  for  the  symmetric  arch. 

For  cos,/?  =0.9987  span  1  =  174.4  ft.,  and  2t/ Wc  =  1482.6  from  Table  79M,  then 

etl  =  -0.0752  ft.      and 
_0.0752X3,775,720 


from  which  the  temperature  stresses  for  any  section  are  readily  obtained  by  computing 
the  moments  about  the  two  kernel  points  of  the  section. 

A  complete  investigation  of  the  stresses  in  aiiy  arch  would  then  consist  in  finding 
the  kernel  moments  due  to  the  dead  loads,  the  kernel  moments  due  to  the  train  of  con- 
centrated live  loads,  Fig.  79o,  making  proper  allowance  for  impact;  and  finally  the  kernel 
moments  due  to  temperature  effects;  all  for  each  of  the  five  critical  sections.  These 
moments  for  the  same  section  are  algebraically  combined  into  the  total  Me  and  Mt-  moments 
and  from  them  and  Eqs.  (76c)  the  maximum  stresses  are  found. 


358 


KINETIC  THEORY  OF  ENGINEERING  STRUCTURES 


CHAT.  XV 


TABLE  79E 

COMPUTATION  OF  MOMENTS   AND  STRESSES  ON  THE  CRITICAL  SECTIONS 
FOR  LOADS  Q+P— UNSYMMETRIC   ARCH 


SECTION  A. 

SECTION  m. 

SECTION  n. 

Point. 

Loads 

Q  +  P 

fie 

m 

Me 

Mi 

i?e 

ft 

Me 

Mi 

to 

w 

Me 

*< 

cu.ft. 

ft. 

ft. 

ft. 

ft. 

ft. 

ft. 

1 

367.4 

-    3.12 

-   5.99 

-1146.3 

-2200.7 

0.25 

0.39 

91.9 

143.3 

+  0  05 

+  0.03 

+      18.4 

+      11.0 

3 

276.1 

-   9.61 

-12.53 

-2653.3 

-3459.5 

2.17 

2.07 

599.1 

671.5 

-0.42 

-0.62 

-    116.0 

-    171.2 

5 

223.4 

-10.40 

-13.43 

-2323.4 

-3000.3 

6.01 

5.49 

1342.6 

1226.5 

-1.08 

-1.52 

-   241.3 

-    339.6 

7 

182.7 

-   7.06 

-10.14 

-1289.9 

-1852.6 

7.98 

6.03 

1457.9 

1101.7 

^0.69 

-1.53 

-    125.1 

-   279.5 

9 

168.0 

-    1.65 

-4.70 

-   277.2 

-   789.6 

3.08 

0.84 

517.4 

141.1 

4-1.10 

-0.05 

+   184.8 

-       8.4 

11 

288.3 

+  4.72 

+    1.87 

+  1360.8 

+   539.1 

-0.38 

-2.48 

-    109.5 

-   715.0 

5.04 

+  3.65 

1453.0 

+  1052.3 

13 

288.3 

9.79 

7.30 

2822  .  4 

2104.6 

-1.99 

-4.18 

-    573.7 

-1205.1 

6.50 

5.06 

1874  .  1 

1458.  S 

15 

267.3 

12.18 

10.18 

3255.7 

2721  .  1 

-2.39 

-3.97 

-    638.8 

-1051.2 

-1-1.73 

-1-0.42 

-1-   462.4 

+    112.3 

17 

282.0 

11.35 

9.88 

3200  .  7 

2786.2 

-2.02 

—  3.10 

-    569.6 

-   874.2 

-0.36 

-1.40 

-    101.5 

—   394.8 

19 

322.7 

8.86 

8.12 

2859  .  1 

2620  .  3 

-1.23 

-1.88 

-    396.9 

-   606.7 

-1.10 

-1.82 

—   355.0 

—   587.3 

21 

375.4 

5.36 

4.89 

2012.1 

1835.7 

-0.58 

-1.02 

-   217.7 

-   382.9 

-1.10 

-1.54 

-   412.9 

-    578.1 

23 

466.7 

2.16 

2.01 

1008  .  1 

938.1 

-0.17 

-0.19 

-      79.4 

-     88.7 

-0.45 

-0.59 

-    210.0 

-   275.4 

25* 

764.0 

0.42 

0.40 

320.9 

305.6 

-0.01 

0.00 

7.6 

-       0.0 

-0.08 

-0.11 

-      61.1 

-     84.0 

Totals 

4272.3 

D= 

10'.  0 

+  9149.7 

+  2548.0 

D= 

6'.90 

+  1415.7 

-1739.7 

D= 

6'.  00 

+  2369.8 

-     S3.S 

N= 

1980.5 

N= 

1372.0 

A"  = 

1226.9 

v  = 

+  2'.953 

v  = 

-OM19 

v  = 

-f  0'.932 

f— 

-   549.0 

+    152.9 

/= 

-    178.4 

-    219.7 

f  

-    395.0 

-      14.0 

SECTION  in'. 

SECTION  B. 

Point 

Loads. 

Q  +  P 

Tie 

9» 

Me 

Mi 

1e 

fc 

Me 

tfi 

cu  ft. 

ft. 

ft. 

ft. 

ft. 

1 

367.4 

-0.03 

-0.09 

11.0 

-     33.0 

0.52 

0.50 

191.0 

183.7 

3 

276.1 

-0.66 

-0.84 

-    182.2 

-   231.9 

3.83 

3.53 

1057  .  5 

974.6 

5 

223.4 

-1.44 

-2.00 

-   321.7 

-   446.8 

9.58 

8.80 

2140.2 

1965.9 

7 

182.7 

-2.32 

-3.61 

-   323.9 

-   659.5 

14.79 

13.39 

2702.0 

2446.4 

9 

168.0 

-3.17 

-4.62 

-   532.6 

-   776.2 

18.80 

16.69 

3158.4 

2803.9 

11 

288.3 

-3.42 

-5.46 

-   986.0 

-  1574  .  1 

19.43 

16.51 

5601.7 

4759.8 

13 

288.3 

-2.09 

-4.37 

-   602.5 

-1259.9 

15.93 

12.33 

4592.6 

3554.7 

15 

267.3 

+  0.22 

-1.80 

+      58.7 

-   481.1 

8:39 

4.24 

2242  .  6 

1133.4 

17 

282.0 

4.71 

+  2.71 

1328.2 

+   764.2 

0.49 

-   3.97 

138.2 

-1119.5 

19 

322.7 

8.66 

7.75 

2794.6 

2500.9 

-   6.67 

-11.34 

-2152.4 

-3659.4 

21 

375.4 

3.72 

3.64 

1396.5 

1365.4 

-10.64 

-15.37 

-3994.3 

-5769.9 

23 

466.7 

+  1.05 

0.97 

+   490.1 

452.7 

-   9.46 

-14.20 

-4415.0 

-  6627  .  1 

25i 

764.0 

-0.05 

0.02 

-      38.  2 

15.3 

-    1.76 

-   6.80 

-1344.6 

-5195.2 

Totals 

4272.3 

D  = 

7'.  20 

+  3070.0 

-   374.0 

D  = 

14'.  7 

+  9917.9 

-4548.7 

N= 

1435.0 

;V  = 

2952.4 

V  — 

+  0    940 

V  — 

+  0'  .  909 

/= 

-   355.3 

43.3 

/- 

-    274.7 

-    126.  C 

Eqs.  (76c, 


3 

-Jj{ 


D2 


v  is  positive  when  measured 


from  the  axis  toward  the  extrados.  All  loads  in  above  table  are  expressed  in  cubic  feet  of  masonry 
and  the  stresses  are  cubic  feet  per  square  foot.  Cubic  feet  per  square  foot  X  0.972  =  pounds  per  square 
inch. 


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ASIMONT.     Hauptspannung  und  Sekundaerspannung.     Zeitschr.  fuer  Baukunde,  1880. 
MANDERLA.     Die  Berechnung  der  Sekundaerspannungen,  etc.,    Allgemeine  Bauzeitung,  1880. 
MUELLER-BRESLAU.     Ueber  Biegungsspannungen  in  Fachwerken.     Allgemeine   Bauzeitung,    1885. 


BIBLIOGRAPHY  361 

LANDSBERG.  Ebene  Fachwerksysteme  mit  festen  Knotenpunkten,  etc.  Centralb.  der'  Bauver- 
waltung,  1885. 

LANUSBERG.  Beitrag  zur  Theorie  des  Fachwerks.  Zeitschr.  des  Arch.-  und  Ing.-Vereins,  Han- 
nover, 1885  and  1886. 

MUELLER-BRESLAU.  Zur  Theorie  der  Biegungsspannungen  in  Fachwerktraegern.  Zeitschr.  des 
Arch.-  und  Ing.-Vereins,  Hannover,  1886. 

WINKLER.     Querkonstruktionen.     Zeitschr.  des  Arch.-  und  Ing.-V.,  Hannover,  1886. 

Handbuch  der  Ingenieurwissenschaften,  Vol.  II,  1890. 

MOHR.     Die  Berechnung  der  Fachwerke  mit  starren  Knotenverbindungen.     Civiling.,  1891  and  1892. 

ENGESSER.     Die  Zusatzkraefte  und  Nebenspannungen  eiserner  Fachwerke,  1892-3. 

HIROI.     Statically  Indeterminate  Bridge  Stresses,  1905. 

MEHRTENS.     Statik  der  Baukonstruktionen,  Vol.  Ill,  1905. 

MOHR.     Abhandlungen  aus  dem  Gebiete  der  technischen  Mechanik,  1906. 

GRIMM.     Secondary  Stresses  in  Bridge  Trusses,  1908. 

SPECIAL  TREATISES  ON  ARCHES 

LAME  and  CLAPEYRON.  Journal  des  Voies  de  Communication,  1826.  This  contains  the  first  applica- 
tion of  force  and  equilibrium  polygons  to  fixed  arches. 

PONCELET.     Memorial  de  1'officier  du  genie,  1835. 

GERSTNER.     Handbuch  der  Mechanik,  1831.     Line  of  thrust  method. 

MOSELEY.     Phil.  Mag.,  1833.     Discussion  of  most  probable  position  of  line  of  thrust. 

HAGEN.     Ueber  Form  und  Staerke  gewoelbter  Bogen,  1844. 

SCHWEDLER.     Theorie  der  Stuetzlinie.     Zeitschr.  fuer  Bauwesen,  1859. 

CULMANN.     Die  graphische  Statik,  1866. 

FRAENKEL.     Berechnung  eiserner  Bogenbruecken.     Civiling.,  1867,  1875. 

MOHR.  Beitrag  zur  Theorie  der  elastischen  Bogentraeger.  Zeitschr.  des  Arch.-  und  Ing.-Ver.. 
Hannover,  1870,  1874,  1881. 

WINKLER.  Beitrag  zur  Theorie  der  elastischen  Bogentraeger.  Zeitschr.  des  Arch.-  und  Ing.-Ver., 
Hannover,  1879. 

WEYRAUCH.     Theorie  der  elastischen  Bogentraeger.     Zeitschr.  fuer  Baukunde,  1878. 

ENGESSER.     Theorie  und  Berechnung  der  Bogenfachwerktraeger,  1880. 

MUELLER-BRESLAU.    Theorie  und  Berechnung  der  eiserner  Bogenbruecken,  1880. 

W.  RITTER.     Der  elastische  Bogen  berechnet  mit  Hilfe  der  graphischen  Statik,  1886. 

MUELLER-BRESLAU.     Graphische  Statik  der  Baukonstruktionen,  1892,  1907. 

WEYRAUCH.     Die  elastischen  Bogentraeger,  1897. 

TOLKMITT.     Leitfaden  fuer  das  Entwerfen  gewoelbter  Bruecken,  1895. 

MEHRTENS.     Statik  der  Baukonstruktionen,  3  vols.,  1903-5. 

Handbuch  der  Ingenieurwissenschaften,  Vol.  I,  1904. 


INDEX 


Abutment  displacements,  fixed  framed  arches .   188 
for  general  case  of  re- 
dundancy    208 

in  two-hinged  arches, 

156,  165,  168 

stresses  due  to 32 

Additional  stresses  due  to  dynamic  influences .   256 

Arches,  bibliography  of  special  treatises  on 361 

fixed  framed 178,  197 

fixed  solid  web  or  masonry 298 

three-hinged 72,. 78 

two-hinged 153,    166,  176 

without  hinges 178,  298 

B 

Beams,  deflection  of 117,  118,  123,  125,  126 

work  of  deformation  for 41 

Betti's  law 28 

Bibliography  of  treatises,  etc 359 


Cantilever  bridges  by  influence  lines 67 

Castigliano's  law,  derivative  of  work  equation .  30,  31 

Centrifugal  force  due  to  curved  track 260 

Changes  in  the  angles  of  a  triangle  due  to 

stress 107 

Choice  of  the  redundant  conditions 202 

Clapeyron's  law 2,  14 

Column  formulae 270 

Combination  of  stresses  as  a  basis  for  design- 
ing    268 

Continuous  girder  over  four  supports 130 

three  supports.  129,  143,  148 
Critical    loading    for    max.    moments,    direct 

loading 59 

Critical   loading  for  max.   moments,   indirect 

loading 60 

Critical  loading  for  max.  shear,  direct  loading. .     62 

,  indirect  loading    62 


•     D 

PAGE 

Dead-load  stresses  in  determinate  structures  .  .  210 
Definitions  of  terms  used  throughout  this  work .     vii 

Deflection  influence  lines 49 

of  a  solid  wreb  beam 117 

polygons  according  to  Prof.  Laud  .  .  107 
polygons  according  to  Prof.  Mohr, 

100,  104 
polygons  for  determinate  structures.     99 

polygons  for  the  loaded  chord 104 

Deformations 11 

Deformation,  work  due  to 1,  29 

Derivative  of  the  work  equation;   Castigliano's 

law...  .\ 30,  31 

Designing,  combination  of  stresses  as  a  basis 

for 268 

Determinate  structures,  dead  load  stresses  in ..  210 
displacements    of 

points 18 

displacement  influence 

lines  for 122 

influence  lines  for.  46,  65 

stresses  in 210,  218 

Direct  and  indirect  loading 47,  48,  50 

Displacement     influence     lines,     determinate 

structures 122 

for  a  cantilever.  .   124 
for  a  simple  beam 
ortruss.123,125,126 

Displacements,  horizontal 114 

of  points  for  determinate  struc- 
tures      18 

Distortion  of  a  statically  determinate  frame  by 


graphics. 


87 


Distortions  due  to  changes  in  the  lengths  of 

members 87 

Double  intersection  trusses  by  influence  lines .  .  65 

Dynamic  impact 262 

work  of  deformation  due  to. ...  43 

influences ' 256 

363 


364 


INDEX 


E 

PAGE 

Eccentric  connections,  stresses  produced  by. . .  255 

Effects  due  to  unusual  loads 266 

Effect  of  shop  lengths  on  determinate  structures  256 
indeterminate     struc- 
tures    208 

Elastic  deformations 11 

Elastic  loads  w  in  terms  of  angle  changes.   109,  111 

Equation  of  an  influence  line 47 

Equations,  solution  of  simultaneous 295 

Externally  indeterminate,  definition  of 6 

External  redundant,  influence  line  for  one .  128,  140 

F 

Fatigue  of  material 265 

Features  in  design  intended  to  diminish  secon- 
dary stresses 267 

Fixed  framed  arches 178 

effect   of   abutment   dis- 
placements    188 

example 189 

influence  lines  for  the  re- 

dundants 184 

location  of  the  coordinate 

axes 182 

resultant  polygon  for ....  187 
stress  influence  lines  for..  185 
temperature  stresses  in  . .  188 

Fixed  masonry  arches 298 

coordinate     axes      and 
elastic  loads,  314,  331,  346 

critical  sections 322 

determination     of    the 

redundants 309 

example  of  symmetric. .   326 
example  of  unsymmet- 

ric 346 

influence   lines   for  the 

redundants.  317,  333,  351 
maximum  stresses, 

322,  344,  357 
modern  methods  of  con- 
struction    301 

preliminary  design  for, 

304,  326 
resultant  polygons, 

324,  335,  357 
stresses  on  any  normal 

section 320,  344 

temperature  stresses  in, 

318,  345,  357 


Framed  structures,  theorems,  laws  and  formulae 
for 11 

G 

General  case  of  redundancy 203 

Girder  continuous  over  four  supports 130 

three  supports.  129,  143,  148 
fixed  at  one  end,  supported  at  the  other .    140 

Glossary  of  terms vii 

Graphostatics,  bibliography 359 

H 
Horizontal  displacements 114 


Impact  due  to  dynamic  effects  ............... 

formulae  ........................... 

Indeterminate,  externally  ................... 

internally  .................... 

Indeterminate  straight  beams,  work  of  defor- 
mation .................... 

structures  by  Maxwell's  law.  ... 

by  Mohr's  work  equa- 
tion .............. 

effect    of    abutment 
displacements.    32, 
effect  of  shop  lengths 
effect  of  temperature 
influence  lines  for, 

127, 

influence  line  for  one 
redundant  ....... 

stress  influence  lines 
for  ......   137,  139, 

Influence  line,  equation  for  .................. 

defined  ....................... 

for  cantilever  bridges  .......... 

deflections  ................ 

determinate  structures, 

46,  65, 
direct  loading  .............. 

displacements  of  determinate 
structures  ............... 

double  intersection  trusses  ... 
indeterminate  structures.  127, 
indirect  loading  ............ 

moments  ................  49 

one  external  redundant.  128, 
one  internal  redundant  ..... 

one  redundant  condition.  .  .  . 

shear  ...................  43 


262 

264 

6 


41 
24 

19 

208 
208 
206 

140 
127 

140 

47 
46 
67 
49 

218 

48 

122 
65 
140 
50 
,  50 
140 
128 
127 
,  50 


INDEX 


365 


PAGE      I 

Influence  line  for  skew  plate  girder  on  curved 

double  track 80 

stresses  in  truss  members  ...     52 

reactions 48,  50 

redundants  in  fixed  masonry 

arches 317,  333 

three-hinged  arches 72 

three-hinged  masonry  arches     78 
three-hinged    solid    web 

arches 78 

trusses    with    subdivided 

panels 83 

two  redundant  conditions.  . .   130 
Internal  redundant,  influence  lines  for  one ....   128 

Internally  indeterminate 6,  7 

Introduction 1 

Isotropic  bodies,  theorems,  laws  and  formulae 
for 34 

K 

Kernel  moment  influence  areas  for  three- 
hinged  solid  web  arch 79 

Kernel  moment  influence  areas  for  two-hinged 
solid  web  arches 159 


Lateral  trusses,  stresses  due  to  wind,  etc . .   258,  261 

vibrations 260 

Least  work,  theorem  of 29 

Live  load  stresses 218,  219 

author's  method 219 

Load  divide  for  a  truss 51 

Loading,  direct  and  indirect .  47,  48,  50 

M 

Masonry  arches,  see  under  Arches. 

Maxwell's  theorem  of  reciprocal  displacements .     27 

Menabrea's  law  of  least  work 29,  31 

Mitering  lock  gates 271 

example 279 

theory  of  the  analysis  ....   273 

Mohr's  rotation  diagram 89 

work  equations 16,  17 

Moment  influence  lines 49,  50 

Moving  load  stresses 218,  219 

Multiple  redundancy 132,  137 

simplification  of  influence 
lines  for 132,  135 

N 
Nature  of  secondary  stresses 226 


Plate  girder  continuous  over  four  supports  ....    130 
over  three  supports, 

129,  143,  148 

deflection  of 117 

fixed  at  one  end  and  supported  at 

the  other 140 

Positions  of  a  moving  load  for  max.  moments. .  .      57 

for  max.  shears 62 

Preface v 

Principle  of  virtual  velocities 2,  16 

work 1,  15 

R 

Reaction  conditions 5 

influence  lines 48,  50- 

summation  influence  lines 54 

Redundancy,  simplification   of   influence   lines 

for 132,  135 

solution  of  the  general  case  of ...   203- 
stress   influence   lines   for  struc- 
tures involving  same .   137,139,140 

Redundant  conditions. 5 

influence  lines  for  one  ...   128 
influence  lines  for  two . .  .    130 

on  the  choice  of 202 

Ritter's  method  of  moments 211 

Rotation  diagram,  Mohr's 89 

of  a  rigid  frame  about  a  fixed  point. .     89 

S 

Secondary    and    additional    stresses,    bibliog- 
raphy    360 

Secondary  stresses 226 

as  effected   by  certain  de- 
signs    267 

concluding  remarks  on 267 

due  to  riveted  connections, 

227,  235 

due  to  various  causes 251 

in  members  due  to  eccentric 

connections 255 

in  members  due  to  fheir  own 

weight 251 

in  pin-connected  structures.  253 
in  riveted  cross-frames,  load 

effects 244 

in  riveted  cross-frames,  wind 

effects 247 

the  nature  of 226 

Shear  influence  lines 49,  50 

Shearing  stress,  work  of  deformation  due  to. .     38 


366 


INDEX 


Simplification  of  influence  lines  for  multiple 

redundancy 132,  135 

Skew  plate  girder  on  curved  double    track  by 

influence  lines 80 

Solid  web  beam,  deflection  of 117 

Solution  of  simultaneous  equations 295 

Special  applications  of  influence  lines  to  inde- 
terminate structures 140 

Statically  determinate  frame,  distortion  of  by 

graphics 87 

structures,      deflection 

polygons  for 99 

indeterminate  structures 140,  194 

influence  lines  for.    140 
methods    of    pre- 
liminary designing  194 

Stresses  by  Ritter's  method  of  moments 211 

the  method  of  influence  lines 218 

stress  diagrams 213 

Stresses,  combination  of  as  a  basis  for  designing  268 

due  to  abutment  displacements 32 

in  statically  determinate  structures, 

210,  218 

on  any  arch  section 157,  320 

Stress  influence  lines  for    indeterminate    struc- 
tures     137,  139,  140 

for  truss  members 52 

Structural  redundancy,  tests  for 6,7 

Summation  influence  lines  for  reactions . .  54 


Temperature  effects  on  determinate  structures .   256 
indeterminate       struc- 
tures     206 

stresses   follow    Menebrea's    and 

Castigliano's  laws 31 

fixed  framed  arches 188 

fixed  masonry  arches, 

318,  345 

girder  on  three  supports .  146 
truss  on  three  supports ..  152 
two-hinged  arches, 

156,  165,  167 

Tests  for  structural  redundancy 7 

Theorem  of  least  work 29 

Theorems  relating  to  work  of  deformation ....     29 

Theory  of  elasticity,  bibliography 359 

Three-hinged  framed  arches  by  influence  lines .     72 
solid  web  and  masonry  arches  by 
influence  lines. .  78 


Tractive  forces,  effect  of j 

Trusses  with  subdivided  panels ; 

Truss  on  three  supports,  problem 1 

Two-hinged  arch  with  cantilever  side  spans.  .  .    170 

tension  member 1 

framed  arch,  abutment     displace- 
ments      168,  1 

deflection     for     any 

point 168,  1 

example 166,  1  9 

stress*  influence  areas 

for 168,  1 

temperature    stresses 

in 167, 17" 

with  column  supports  I.'/ 
Xa  influence  line  for, 

166,  1 

solid  web  'arch,  abutment  displace- 
ments     156,  ] 

example 153,  1 

kernel  moment  in- 
fluence areas . . . .    ] 
temperature  stresses, 
156,  ] 

Xa  influence  line  for, 
154, 


U 


Unusual  load  effects. 


V 

Virtual  velocities , 2, 

work,  principle  of 1 , 

W 

Williot  displacement  diagrams 

Williot-Mohr  diagrams 

Wind  pressure 2 

stresses  in  cross-frames 

in  main  and  lateral  trusses 2 

Work  equations  for  any  frame 14 

for  isotropic  solids 

of  deformation 1 

due  to  dynamic  impact..  . 
due  to  shearing  stress .... 

due  to  moments 

for     any     indeterminat  i 

straight  beam 

theorems  relating  to 


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